Idea Transcript
4
4
Because, (3c − d) = [3c + (−d)] , write the series 4
for (a + b) , replacing a with 3c and b with −d.
10-5 The Binomial Theorem Use Pascal's triangle to expand each binomial.
1. (2 + x)
4
8. (m – a)5 SOLUTION:
SOLUTION:
5
5
Because, (m − a) = [m + (−a)] , write the series for 5 (a + b) , replacing a with m and b with −a.
2. (n + m)5
9. (a – b)3
SOLUTION:
SOLUTION: 3
3
Because, (a − b) = [a + (−b)] , write the series for 3
3. (4a – b)3
(a + b) , replacing a with a and b with −b.
SOLUTION: 3
3
Because, (4a − b) = [4a + (−b)] , write the series 3
for (a + b) , replacing a with 4a and b with −b.
10. (3p – 2q)4 SOLUTION: 4
4
Because, (3p − 2q) = [3p + (−2q)] , write the 4 series for (a + b) , replacing a with 3p and b with −2q.
4. (x + y)6 SOLUTION:
Find the coefficient of the indicated term in each expansion.
5. (3x + 2y)
7
11. (x – 2)10, 5th term
SOLUTION:
SOLUTION: 10
The fifth term in the expansion of (x – 2) 6
6. (n – 4)
found by evaluating a or 4.
SOLUTION: 6
can be
n –r r
b for n = 10 and r = 5 – 1
6
Because, (n − 4) = [n + (−4)] , write the series for 6 (a + b) , replacing a with n and b with −4.
10
For (x – 2)
n
to have the form (a + b) , let a = x and n–
b = –2. The coefficient of the term containing a r r n b in the expansion of (a + b) is given by nCr. So,
7. (3c – d)4
to find the binomial coefficient of the term containing 6 4 10 a b in the expansion of (a + b) , evaluate nCr for
SOLUTION: 4
4
Because, (3c − d) = [3c + (−d)] , write the series
n = 10 and r = 4.
4
for (a + b) , replacing a with 3c and b with −d.
8. (m – a)5 eSolutions Manual - Powered by Cognero SOLUTION:
5
Page 1
5
Because, (m − a) = [m + (−a)] , write the series for 5
Because, (3p − 2q) = [3p + (−2q)] , write the 4 series for (a + b) , replacing a with 3p and b with −2q.
Therefore, the coefficient of the fifth term in the 10 expansion of (x – 2) is 3360.
10-5 The Binomial Theorem
12. (4m + 1)8, 3rd term
Find the coefficient of the indicated term in each expansion.
SOLUTION:
11. (x – 2)10, 5th term
8
The third term in the expansion of (4m + 1) can be
SOLUTION: 10
The fifth term in the expansion of (x – 2) found by evaluating a or 4.
found by evaluating a or 2.
can be
n –r r
n –r r
b for n = 8 and r = 3 – 1
b for n = 10 and r = 5 – 1 8
n
For (4m + 1) to have the form (a + b) , let a = 4m n
10
For (x – 2)
n
to have the form (a + b) , let a = x and n–
b = –2. The coefficient of the term containing a r r n b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 6 4 10 a b in the expansion of (a + b) , evaluate nCr for
and b = 1. The coefficient of the term containing a –r r n b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 6 2 8 a b in the expansion of (a + b) , evaluate nCr for n = 8 and r = 2.
n = 10 and r = 4.
6 2
Thus, the binomial coefficient of the a b term in (a 8
6 4
Thus, the binomial coefficient of the a b term in (a 10
+ b) is 210. Substitute x for a and –2 for b to find the coefficient for the fifth term in the original binomial expansion.
Therefore, the coefficient of the third term in the 8 expansion of (4m + 1) is 114,688.
Therefore, the coefficient of the fifth term in the 10 expansion of (x – 2) is 3360.
13. (x + 3y)10, 8th term SOLUTION:
12. (4m + 1)8, 3rd term
The eighth term in the expansion of (x + 3y)
SOLUTION: 8
The third term in the expansion of (4m + 1) can be found by evaluating a or 2.
+ b) is 28. Substitute 4m for a and 1 for b to find the coefficient for the third term in the original binomial expansion.
be found by evaluating a – 1 or 7.
10
can
n –r r
b for n = 10 and r = 8
n –r r
b for n = 8 and r = 3 – 1 For (x + 3y)
10
n
to have the form (a + b) , let a = x n
8
n
For (4m + 1) to have the form (a + b) , let a = 4m n
and b = 1. The coefficient of the term containing a –r r n b in the expansion of (a + b) is given by nCr. So, eSolutions Manual - Powered by Cognero
to find the binomial coefficient of the term containing 6 2 8 a b in the expansion of (a + b) , evaluate nCr for n
and b = 3y. The coefficient of the term containing a –r r n b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 3 7 10 a b in the expansion of (a + b) , evaluate nCr for n = 10 and r = 7.
Page 2
the coefficient of the third term in the 10-5Therefore, The Binomial Theorem 8
Therefore, the coefficient of the eighth term in the 10 expansion of (x + 3y) is 262,440.
expansion of (4m + 1) is 114,688.
13. (x + 3y)10, 8th term
14. (2c – d)12, 6th term
SOLUTION:
SOLUTION:
The eighth term in the expansion of (x + 3y) be found by evaluating a – 1 or 7.
For (x + 3y)
10
10
The sixth term in the expansion of (2c – d)
can
n –r r
b for n = 10 and r = 8
found by evaluating a or 5.
n
n
and b = 3y. The coefficient of the term containing a –r r n b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 3 7 10 a b in the expansion of (a + b) , evaluate nCr for n = 10 and r = 7.
3 7
Thus, the binomial coefficient of the a b term in (a 10
+ b) is 120. Substitute x for a and 3y for b to find the coefficient for the eighth term in the original binomial expansion.
Therefore, the coefficient of the eighth term in the 10 expansion of (x + 3y) is 262,440.
n –r r
n
7 5
7 5
Thus, the binomial coefficient of the a b term in (a 12 + b) is 792. Substitute 2c for a and (–d) for b to find the coefficient for the sixth term in the original binomial expansion.
Therefore, the coefficient of the sixth term in the expansion of (2c – d)
SOLUTION:
found by evaluating a or 5.
n –r r
b for n = 12 and r = 6 – 1
12
n
For (2c – d) to have the form (a + b) , let a = 2c and b = (–d). The coefficient of the term containing n –r r
n
a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term 7 5
12
containing a b in the expansion of (a + b) , eSolutions ManualnC - Powered by12 Cognero evaluate and r = 5. r for n =
12
containing a b in the expansion of (a + b) , evaluate nCr for n = 12 and r = 5.
SOLUTION: can be
n
a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term
15. (a + b)8, 4th term 12
b for n = 12 and r = 6 – 1
For (2c – d) to have the form (a + b) , let a = 2c and b = (–d). The coefficient of the term containing
14. (2c – d)12, 6th term
The sixth term in the expansion of (2c – d)
can be
n –r r
12
to have the form (a + b) , let a = x
12
12
is −101,376.
8
The fourth term in the expansion of (a + b) can be n –r r found by evaluating x y for n = 8 and r = 4 – 1 or 3.
8
n
For (a + b) to have the form (x + y) , let x = a and n –r r y = b. The coefficient of the term containing x y n
in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 5 3
8
x y in the expansion of (x + y) , evaluate nCr for n Page 3 = 8 and r = 3.
Therefore, the coefficient of the sixth term in the
Therefore, the coefficient of the fourth term in the
expansion of (2c – d)
expansion of (a + b) is 56.
10-5 The Binomial Theorem 12
8
is −101,376.
15. (a + b)8, 4th term
16. (2a + 3b)10, 5th term
SOLUTION:
SOLUTION: 8
The fourth term in the expansion of (a + b) can be n –r r found by evaluating x y for n = 8 and r = 4 – 1 or 3.
8
n
For (a + b) to have the form (x + y) , let x = a and n –r r y = b. The coefficient of the term containing x y n
in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 5 3
8
x y in the expansion of (x + y) , evaluate nCr for n = 8 and r = 3.
5 3
10
The fifth term in the expansion of (2a + 3b) can be n –r r found by evaluating x y for n = 10 and r = 5 – 1 or 4.
10
n
For (2a + 3b) to have the form (x + y) , let x = 2a n and y = 3b. The coefficient of the term containing x – r
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 6 4
10
x y in the expansion of (x + y) , evaluate nCr for n = 10 and r = 4.
6 4
Thus, the binomial coefficient of the x y term in (x 8 + y) is 56. Substitute a for x and b for y to find the coefficient for the fourth term in the original binomial expansion.
Thus, the binomial coefficient of the x y term in (x 10 + y) is 210. Substitute 2a for x and 3b for y to find the coefficient for the fifth term in the original binomial expansion.
Therefore, the coefficient of the fourth term in the
Therefore, the coefficient of the fifth term in the
8
expansion of (a + b) is 56.
expansion of (2a + 3b)
16. (2a + 3b)10, 5th term
10
is 1,088,640.
17. (x – y)9, 6th term
SOLUTION:
SOLUTION: 10
The fifth term in the expansion of (2a + 3b) can be n –r r found by evaluating x y for n = 10 and r = 5 – 1 or 4.
10
n
9
The sixth term in the expansion of (x – y) can be n –r r found by evaluating a b for n = 9 and r = 6 – 1 or 5.
9
n
For (2a + 3b) to have the form (x + y) , let x = 2a n and y = 3b. The coefficient of the term containing x
For (x – y) to have the form (a + b) , let a = x and n– b = –y. The coefficient of the term containing a
– r
r r
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 6 4
10
x y in the expansion of (x + y) , evaluate nCr for n = 10 and r = 4.
eSolutions Manual - Powered by Cognero
n
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 4 5
9
a b in the expansion of (a + b) , evaluate nCr for n Page 4 = 9 and r = 5.
Therefore, the coefficient of the fifth term in the
Therefore, the coefficient of the sixth term in the
expansion of (2a + 3b)
expansion of (x – y) is −126.
10-5 The Binomial Theorem 10
9
is 1,088,640.
17. (x – y)9, 6th term
18. (x + y)12, 7th term
SOLUTION:
SOLUTION: 9
The sixth term in the expansion of (x – y) can be n –r r found by evaluating a b for n = 9 and r = 6 – 1 or 5.
9
n
For (x – y) to have the form (a + b) , let a = x and n– b = –y. The coefficient of the term containing a r r
n
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 4 5
9
a b in the expansion of (a + b) , evaluate nCr for n = 9 and r = 5.
4 5
12
The seventh term in the expansion of (x + y) can n –r r be found by evaluating a b for n = 12 and r = 7 – 1 or 6.
12
n
For (x + y) to have the form (a + b) , let a = x and n –r r b = y. The coefficient of the term containing a b n
in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 6 6
12
a b in the expansion of (a + b) , evaluate nCr for n = 12 and r = 6.
6 6
Thus, the binomial coefficient of the a b term in (a 9 + b) is 126. Substitute x for a and (–y) for b to find the coefficient for the sixth term in the original binomial expansion.
Thus, the binomial coefficient of the a b term in (a 12 + b) is 924. Substitute x for a and y for b to find the coefficient for the seventh term in the original binomial expansion.
Therefore, the coefficient of the sixth term in the
Therefore, the coefficient of the seventh term in the
9
expansion of (x – y) is −126.
expansion of (x + y)
18. (x + y)12, 7th term
is 924.
19. (x + 2)7, 4th term
SOLUTION:
SOLUTION: 12
The seventh term in the expansion of (x + y) can n –r r be found by evaluating a b for n = 12 and r = 7 – 1 or 6.
12
n
For (x + y) to have the form (a + b) , let a = x and n –r r b = y. The coefficient of the term containing a b n
in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 6 6
12
12
a b in the expansion of (a + b) , evaluate nCr for eSolutions Manual - Powered by Cognero n = 12 and r = 6.
7
The fourth term in the expansion of (x + 2) can be n –r r found by evaluating a b for n = 7 and r = 4 – 1 or 3.
7
n
For (x + 2) to have the form (a + b) , let a = x and n –r r b = 2. The coefficient of the term containing a b n
in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 4 3
7
a b in the expansion of (a + b) , evaluate nCr for n Page 5 = 7 and r = 3.
Therefore, the coefficient of the seventh term in the
Therefore, the coefficient of the fourth term in the
expansion of (x + y)
expansion of (x + 2) is 280.
10-5 The Binomial12Theorem
7
is 924.
19. (x + 2)7, 4th term
20. (a – 3)8, 5th term
SOLUTION:
SOLUTION: 7
The fourth term in the expansion of (x + 2) can be n –r r found by evaluating a b for n = 7 and r = 4 – 1 or 3.
7
n
For (x + 2) to have the form (a + b) , let a = x and n –r r b = 2. The coefficient of the term containing a b n
in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 4 3
7
a b in the expansion of (a + b) , evaluate nCr for n = 7 and r = 3.
4 3
8
The fifth term in the expansion of (a − 3) can be n –r r found by evaluating x y for n = 8 and r = 5 – 1 or 4.
8
n
For (a − 3) to have the form (x + y) , let x = a and n– y = −3. The coefficient of the term containing x r r
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 4 4
8
x y in the expansion of (x + y) , evaluate nCr for n = 8 and r = 4.
4 4
Thus, the binomial coefficient of the a b term in (a 7 + b) is 35. Substitute x for a and 2 for b to find the coefficient for the fourth term in the original binomial expansion.
Thus, the binomial coefficient of the x y term in (x 8 + y) is 70. Substitute a for x and −3 for y to find the coefficient for the fifth term in the original binomial expansion.
Therefore, the coefficient of the fourth term in the
Therefore, the coefficient of the fifth term in the
7
8
expansion of (x + 2) is 280.
expansion of (a − 3) is 5670.
20. (a – 3)8, 5th term
21. (2a + 3b)10, a 6b 4 term
SOLUTION:
SOLUTION: 8
The fifth term in the expansion of (a − 3) can be n –r r found by evaluating x y for n = 8 and r = 5 – 1 or 4.
10
–r r
n
For (a − 3) to have the form (x + y) , let x = a and n– y = −3. The coefficient of the term containing x r r
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 6 4
8
n
For (2a + 3b) to have the form (x + y) , let x = 2a n and y = 3b. The coefficient of the term containing x
10
x y in the expansion of (x + y) , evaluate nCr for n = 10 and r = 4.
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 4 4
8
x y in the expansion of (x + y) , evaluate nCr for n eSolutions Manual - Powered by Cognero = 8 and r = 4.
Page 6
6 4
Therefore, the coefficient of the fifth term in the
Therefore, the coefficient of the a b term in the
expansion of (a − 3) is 5670.
expansion of (2a + 3b)
10-5 The Binomial8 Theorem 21. (2a + 3b)10, a 6b 4 term
10
is 1,088,640.
22. (2x + 3y)9, x6y 3 term
SOLUTION:
SOLUTION:
10
n
9
n
For (2a + 3b) to have the form (x + y) , let x = 2a n and y = 3b. The coefficient of the term containing x
For (2x + 3y) to have the form (a + b) , let a = 2x n and b = 3y. The coefficient of the term containing a
–r r
–r r
n
6 3
9
n
y in the expansion of (x + y) is given by nCr. So, to find the binomial coefficient of the term containing 6 4
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing
10
x y in the expansion of (x + y) , evaluate nCr for n = 10 and r = 4.
a b in the expansion of (a + b) , evaluate nCr for n = 9 and r = 3.
6 4
6 3
Thus, the binomial coefficient of the x y term in (x 10 + y) is 210. Substitute 2a for x and 3b for y to find
Thus, the binomial coefficient of the a b term in (a 9 + b) is 84. Substitute 2x for a and 3y for b to find
6 4
6 3
the coefficient for the a b term in the original binomial expansion.
the coefficient for the x y term in the original binomial expansion.
6 4
6 3
Therefore, the coefficient of the a b term in the expansion of (2a + 3b)
10
Therefore, the coefficient of the x y term in the 9
is 1,088,640.
expansion of (2x + 3y) is 145,152.
22. (2x + 3y)9, x6y 3 term
23.
, 4th term
SOLUTION: 9
n
For (2x + 3y) to have the form (a + b) , let a = 2x n and b = 3y. The coefficient of the term containing a –r r
n
The fourth term in the expansion of
9
found by evaluating a or 3.
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 6 3
SOLUTION:
a b in the expansion of (a + b) , evaluate nCr for n = 9 and r = 3.
For and b =
can be
n –r r
b for n = 7 and r = 4 – 1
n
to have the form (a + b) , let a = x . The coefficient of the term containing a
–r r
n
4 3
7
n
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing eSolutions Manual - Powered coefficient by Cognero Thus, the binomial
6 3
of the a b term in (a 9 + b) is 84. Substitute 2x for a and 3y for b to find 6 3
the coefficient for the x y term in the original
a b in the expansion of (a + b) , evaluate nCr for n Page 7 = 7 and r = 3.
Therefore, the coefficient of the fourth term in the 6 3
the coefficient of the x y 10-5Therefore, The Binomial Theorem 9
term in the
23.
24.
, 4th term
SOLUTION: can be
n –r r
b for n = 7 and r = 4 – 1
n
and b =
, 6th term
. The coefficient of the term containing a
n
can be
The sixth term in the expansion of found by evaluating a or 5.
to have the form (a + b) , let a = x
For
.
SOLUTION:
The fourth term in the expansion of found by evaluating a or 3.
is
expansion of
expansion of (2x + 3y) is 145,152.
n –r r
b for n = 10 and r = 6 – 1
n
For
to have the form (a + b) , let a = x
and
. The coefficient of the term containing
n –r r
n
–r r
n
a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term
4 3
7
containing a b in the expansion of (a + b) , evaluate nCr for n = 10 and r = 5.
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing a b in the expansion of (a + b) , evaluate nCr for n = 7 and r = 3.
4 3
Thus, the binomial coefficient of the a b term in (a 7
+ b) is 35. Substitute x for a and
for b to find the
5 5
10
5 5
Thus, the binomial coefficient of the a b term in (a + b)
10
for b to find
is 252. Substitute x for a and
coefficient for the fourth term in the original binomial expansion.
the coefficient for the sixth term in the original binomial expansion.
Therefore, the coefficient of the fourth term in the
Therefore, the coefficient of the sixth term in the
expansion of
is
expansion of
.
is
.
24.
, 6th term
25. (x + 4y)7, x2y 5 term
SOLUTION:
SOLUTION:
The sixth term in the expansion of
For (x + 4y) to have the form (a + b) , let a = x and n– b = 4y. The coefficient of the term containing a Page 8
7
eSolutions Manual - Powered by Cognero
found by evaluating a or 5.
n –r r
can be
b for n = 10 and r = 6 – 1
r r
n
n
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing
Therefore, the coefficient of the sixth term in the is
expansion of
.
2 5
Therefore, the coefficient of the x y term in the
10-5 The Binomial Theorem
7
expansion of (x + 4y) is 21,504.
25. (x + 4y)7, x2y 5 term
26. (3x + 5y)10, x6y 4 term
SOLUTION:
SOLUTION:
7
n
10
n
For (x + 4y) to have the form (a + b) , let a = x and n– b = 4y. The coefficient of the term containing a
For (3x + 5y) to have the form (a + b) , let a = 3x n and b = 5y. The coefficient of the term containing a
r r
–r r
n
6 4
7
n
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing 2 5
7
a b in the expansion of (a + b) , evaluate nCr for n = 7 and r = 5.
2 5
Thus, the binomial coefficient of the a b term in (a 7 + b) is 21. Substitute x for a and 4y for b to find the 2 5
coefficient for the x y term in the original binomial expansion.
2 5
Therefore, the coefficient of the x y term in the 7
expansion of (x + 4y) is 21,504.
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing a b in the expansion of (a + b) , evaluate nCr for n = 10 and r = 4.
6 4
Thus, the binomial coefficient of the a b term in (a 10 + b) is 210. Substitute 3x for a and 5y for b to find 6 4
the coefficient for the x y term in the original binomial expansion.
6 4
Therefore, the coefficient of the x y term in the expansion of (3x + 5y)
26. (3x + 5y)10, x6y 4 term
10
is 95,681,250.
27. TESTING Alfonso is taking a test that contains a
SOLUTION: 10
n
For (3x + 5y) to have the form (a + b) , let a = 3x n and b = 5y. The coefficient of the term containing a –r r
n
6 4
7
b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term containing a b in the expansion of (a + b) , evaluate nCr for n = 10 and r = 4.
eSolutions Manual - Powered coefficient by Cognero Thus, the binomial
6 4
of the a b term in (a 10 + b) is 210. Substitute 3x for a and 5y for b to find 6 4
the coefficient for the x y term in the original
section of 16 true-false questions. a. How many of the possible sets of answers to these questions have exactly 12 correct answers of false? b. How many of the possible sets of answers to these questions have exactly 8 correct answers of true?
SOLUTION: a. Each question represents a trial, so n = 16. You want to find how many of the possible sets of answers to these questions have exactly 12 correct answers of false, so let x = 12. Find the value of the term nCx.
b. Each question represents a trial, so n = 16. You Page 9 want to find how many of the possible sets of answers to these questions have exactly 8 correct answers of true, so let x = 8. Find the value of the
representative making a sale, so p = 6 4
Therefore, the coefficient of the x y term in the
10-5 The Binomial Theorem 10 expansion of (3x + 5y)
is 95,681,250.
27. TESTING Alfonso is taking a test that contains a section of 16 true-false questions. a. How many of the possible sets of answers to these questions have exactly 12 correct answers of false? b. How many of the possible sets of answers to these questions have exactly 8 correct answers of true?
SOLUTION: a. Each question represents a trial, so n = 16. You want to find how many of the possible sets of answers to these questions have exactly 12 correct answers of false, so let x = 12. Find the value of the term nCx.
and q =
.
Each appointment represents a trial, so n = 12. You want to find the probability that the sales representative succeeds 0 times out of those 12 trials, so let x = 0. To find this probability, find the x n –x value of the term nCx p q in the expansion of (p n
+ q) .
The probability that the sales representative makes no sales this week is 6.87%. b. A success in this situation is the sales representative making a sale, so p = or
and q = 1 −
. Each appointment represents a trial, so n =
12. You want to find the probability that the sales representative succeeds 3 times out of those 12 trials, so let x = 3. To find this probability, find the x n –x
value of the term nCx p q
b. Each question represents a trial, so n = 16. You want to find how many of the possible sets of answers to these questions have exactly 8 correct answers of true, so let x = 8. Find the value of the term nCx.
n
+ q) .
The probability that the sales representative makes three sales this week is 23.62%. c. A success in this situation is the sales
28. BUSINESS The probability of a certain sales representative successfully making a sale is
in the expansion of (p
. The
sales representative has 12 appointments this week. a. Find the probability that the sales representative makes no sales this week. b. What is the probability that the sales representative makes exactly 3 sales this week? c. Find the probability that the sales representative will make 10 sales this week.
representative making a sale, so p =
and q =
.
Each appointment represents a trial, so n = 12. You want to find the probability that the sales representative succeeds 10 times out of those 12 trials, so let x = 10. To find this probability, find the x n –x value of the term nCx p q in the expansion of (p n
+ q) .
SOLUTION: a. A success in this situation is the sales representative making a sale, so p =
and q =
.
Each appointment represents a trial, so n = 12. You want to find the probability that the sales representative succeeds 0 times out of those 12 trials, so let x = 0. To find this probability, find the x n –x value of the term nC p q in the expansion of (p eSolutions Manual - Powered byx Cognero n
+ q) .
The probability that the sales representative makes ten sales this week is 0.0004%.
29. BIOLOGY Refer to the beginning of the lesson. Assume that the zoo workers expect 30 gibbonPage 10 offspring this year. a. What is the probability that there will be no male
p=
probability that the sales representative makes 10-5The The Binomial Theorem ten sales this week is 0.0004%.
and q =
. Each offspring represents a trial,
so n = 30. You want to find the probability that the offspring succeeds 23 times out of those 30 trials, so let x = 23. To find this probability, find the value of x n –x
the term nCx p q
n
in the expansion of (p + q) .
29. BIOLOGY Refer to the beginning of the lesson. Assume that the zoo workers expect 30 gibbon offspring this year. a. What is the probability that there will be no male gibbon offspring this year? b. Find the probability that there will be exactly 2 male gibbon offspring this year. c. What is the probability that there will be 23 female gibbon offspring this year?
SOLUTION: a. A success in this situation is a male offspring, so p =
and q =
The probability that there are 23 female offspring is 15.38%.
30. BOWLING Carol averages 2 strikes every 10 frames. What is the probability that Carol will get exactly 4 strikes in the next 10 frames?
. Each offspring represents a trial,
so n = 30. You want to find the probability that the offspring succeeds 0 times out of those 30 trials, so let x = 0. To find this probability, find the value of the x n –x
term nCx p q
n
in the expansion of (p + q) .
SOLUTION: A success in this situation is a strike, so p = =
The probability that there are no male offspring this year is 0.12%. b. A success in this situation is a male offspring, so p =
and q =
. Each offspring represents a trial,
and q
. Each frame represents a trial, so n = 10. You
want to find the probability that Carol succeeds 4 times out of those 10 trials, so let x = 4. To find this x n –x
probability, find the value of the term nCx p q
in
n
the expansion of (p + q) . so n = 30. You want to find the probability that the offspring succeeds 2 times out of those 30 trials, so let x = 2. To find this probability, find the value of the x n –x
term nCx p q
n
in the expansion of (p + q) .
The probability that Carol gets 4 strikes during her next ten frames is 8.8%.
The probability that there are two male offspring this year is 3.37%. c. A success in this situation is a female offspring, so p=
and q =
. Each offspring represents a trial,
Use the Binomial Theorem to expand each binomial. 31. (4t + 3)5
SOLUTION: 5
Apply the Binomial Theorem to expand (a + b) , where a = 4t and b = 3.
so n = 30. You want to find the probability that the offspring succeeds 23 times out of those 30 trials, so let x = 23. To find this probability, find the value of x n –x
the term nCx p q
n
in the expansion of (p + q) .
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32. (8 – 5y)3
Page 11
SOLUTION: 3
Apply the Binomial Theorem to expand (a + b) ,
Apply the Binomial Theorem to expand (a + b) , where a = 4t and b = 3.
10-5 The Binomial Theorem 32. (8 – 5y)3
8
Apply the Binomial Theorem to expand (a + b) , 2 where a = a and b = −2b.
37. (7c2 + 3d)5
SOLUTION:
SOLUTION: 3
Apply the Binomial Theorem to expand (a + b) , where a = 8 and b = −5y.
33. (2m – n)
SOLUTION:
5
Apply the Binomial Theorem to expand (a + b) , 2
where a = 7c and b = 3d.
38. (2w – 4x3)7
6
SOLUTION:
SOLUTION:
7
6
Apply the Binomial Theorem to expand (a + b) , where a = 2m and b = −n.
Apply the Binomial Theorem to expand (a + b) , 3 where a = 2w and b = −4x .
Represent the expansion of each expression using sigma notation.
34. (9h + 2j )4
39. (2q + 3)15
SOLUTION: 4
Apply the Binomial Theorem to expand (a + b) , where a = 9h and b = 2j .
SOLUTION: Apply the Binomial Theorem to represent the 15 expansion of (a + b) using sigma notation, where a = 2q and b = 3.
35. (3p + q)7 SOLUTION: 7
Apply the Binomial Theorem to expand (a + b) , where a = 3p and b = q.
40. (m – 8n)25 SOLUTION: Apply the Binomial Theorem to represent the 25 expansion of (a + b) using sigma notation, where a = m and b = −8n.
36. (a 2 – 2b)8 SOLUTION: 8
Apply the Binomial Theorem to expand (a + b) , 2 where a = a and b = −2b.
41. (11x + y)31 SOLUTION: Apply the Binomial Theorem to represent the 31 expansion of (a + b) using sigma notation, where a = 11x and b = y.
37. (7c2 + 3d)5 SOLUTION: 5
Apply the Binomial Theorem to expand (a + b) , 2
where a = 7c and b = 3d.
42. (4a + 7b)19 eSolutions Manual - Powered by Cognero
3 7
38. (2w – 4x )
SOLUTION:
SOLUTION:
Page 12
Apply the Binomial Theorem to represent the 19 expansion of (x + y) using sigma notation, where x
expansion of (a + b) = 11x and b = y.
31
using sigma notation, where a
10-5 The Binomial Theorem 42. (4a + 7b)19
SOLUTION: a. Each opening of a treasure chest represents a trial, so n = 15. You want to find how many different ways is it possible to open the chest and find gold coins exactly 9 times, so let x = 9. Find the value of the term nCx.
SOLUTION: Apply the Binomial Theorem to represent the 19 expansion of (x + y) using sigma notation, where x = 4a and y = 7b. b. A success in this situation is finding gold coins, so p=
43. SOLUTION: Apply the Binomial Theorem to represent the expansion of (a + b) = 3f and b = −
22
using sigma notation, where a
and q =
. Each opening of a treasure chest
represents a trial, so n = 15. You want to find the probability of finding gold more than 12 times out of those 15 trials, so find the sum of x = 13, x = 14, and x = 15. To find these probabilities, find the value of x n –x n the terms nCx p q in the expansions of (p + q) .
g.
44. SOLUTION: Apply the Binomial Theorem to represent the 22 expansion of (a + b) using sigma notation, where and b = 5t.
45. COMPUTER GAMES In a computer game, when a treasure chest is opened, it contains either gold coins or rocks. The probability that it contains gold coins is
.
a. The treasure chest is opened 15 times per game. In one game, how many different ways is it possible to open the chest and find gold coins exactly 9 times? b. What is the probability that a person playing the game will find gold in the chest more than 12 times?
SOLUTION: a. Each opening of a treasure chest represents a trial, so n = 15. You want to find how many different ways is it possible to open the chest and find gold coins exactly 9 times, so let x = 9. Find the value of the term nCx.
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The probability of finding gold more than 12 times is 0.2726 + 0.1947 + 0.0649 = 0.5322 = 53.22%.
46. COMMUNITY OUTREACH At a food bank, canned goods are received and distributed to people in the community who are in need. Volunteers check the quality of the food before distribution. The probability that a canned good received at a food bank is distributed to the needy is
.
a. A volunteer checks 30 canned goods per hour. In one hour, how many different ways is it possible to check a canned good and donate it exactly 23 times? b. What is the probability that a volunteer checking 13 canned goods will find themselves throwing outPage items less than 4 times?
SOLUTION:
bank is distributed to the needy is
to find the probability of throwing out an item less than 4 times out of those 30 trials, so find the sum of x = 0, x = 1, x = 2 and x = 3. To find these
.
A volunteer checks 30 canned goods per hour. In 10-5a. The Binomial Theorem one hour, how many different ways is it possible to check a canned good and donate it exactly 23 times? b. What is the probability that a volunteer checking canned goods will find themselves throwing out items less than 4 times?
x n –x
probabilities, find the value of the terms nCx p q n
in the expansions of (p + q) .
SOLUTION: a. Each checking of a canned good represents a trial, so n = 30. You want to find how many different ways it is possible to check the canned goods and donate it exactly 23 times, so let x = 23. Find the value of the term nCx.
b. A success in this situation is throwing out an item, so p =
and q = 1 −
or
. Each checking of a
canned good represents a trial, so n = 30. You want to find the probability of throwing out an item less than 4 times out of those 30 trials, so find the sum of x = 0, x = 1, x = 2 and x = 3. To find these x n –x
probabilities, find the value of the terms nCx p q n
in the expansions of (p + q) .
The probability of throwing out an item less than 4 times is 0.0012 + 0.0093 + 0.0337 + 0.0785 = 0.1227 = 12.27%. Use the Binomial Theorem to expand and simplify each expression.
47. (2d +
)
4
SOLUTION: 4
Apply the Binomial Theorem to expand (a + b) , where a = 2d and b =
48. (
–
)
.
5
SOLUTION: 5
eSolutions Manual - Powered by Cognero
Apply the Binomial Theorem to expand (x + y) , where x =
and y = −
.
Page 14
10-5 The Binomial Theorem 48. (
–
)
Find the coefficient of the indicated term in each expansion.
5
SOLUTION:
9
51. (k –
) , 5th term
5
Apply the Binomial Theorem to expand (x + y) , and y = −
where x =
.
SOLUTION: found by evaluating a or 4.
49. SOLUTION: 5
Apply the Binomial Theorem to expand (a + b) , t.
) can be
n –r r
b for n = 9 and r = 5 – 1
9
n
For (k –
) to have the form (a + b) , let a = k
and b = –
. The coefficient of the term containing
n –r r
where a = 4s and b =
9
The fifth term in the expansion of (k –
n
a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of the term 5 4
9
containing a b in the expansion of (a + b) , evaluate nCr for n = 9 and r = 4.
5 4
50.
Thus, the binomial coefficient of the a b term in (a 9
SOLUTION: 6
Apply the Binomial Theorem to expand (a + b) , where a =
+ b) is 126. Substitute k for a and – for b to find the coefficient for the fifth term in the original binomial expansion.
and b = −3z. Therefore, the coefficient of the fifth term in the 9
expansion of (k –
) is 3150.
10
52. (
+ 2c) , middle term
SOLUTION: Each expansion has n + 1 terms; therefore, this expansion has 11 terms. We want to find the middle term or the sixth term. The sixth term in the Find the coefficient of the indicated term in each expansion.
51. (k –
9
10
expansion of ( + 2c) can be found by n –r r evaluating a b for n = 10 and r = 6 – 1 or 5.
) , 5th term
SOLUTION: eSolutions - Powered Cognero TheManual fifth term in thebyexpansion
found by evaluating a or 4.
n –r r
of (k –
9
) can be
b for n = 9 and r = 5 – 1
For (
+ 2c)
10
n
to have the form (a + b) , letPage a = 15
and b = 2c. The coefficient of the term containing a
n –r r
n
b in the expansion of (a + b) is
Therefore, the coefficient of the fifth term in the
Therefore, the coefficient of the sixth term in the
9 10-5expansion The Binomial Theorem of (k – ) is 3150.
expansion of (
+ 2c)
10
is 32,256
.
10
52. (
+ 2c) , middle term
53.
, 7th term
SOLUTION: SOLUTION:
Each expansion has n + 1 terms; therefore, this expansion has 11 terms. We want to find the middle term or the sixth term. The sixth term in the
The seventh term in the expansion of
10
For (
+ 2c)
10
n –r r
can be found by evaluating a
expansion of ( + 2c) can be found by n –r r evaluating a b for n = 10 and r = 6 – 1 or 5.
b for
n = 11 and r = 7 – 1 or 6.
n
to have the form (a + b) , let a =
For
n
to have the form (a + b) , let a =
and b = 2c. The coefficient of the term n –r r
n
containing a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of
p and b = q. The coefficient of the term n –r r
n
containing a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of
5 5
the term containing a b in the expansion of (a + b) 10 , evaluate nCr for n = 10 and r = 5.
5 6
the term containing a b in the expansion of (a + b) 11 , evaluate nCr for n = 11 and r = 6.
5 5
Thus, the binomial coefficient of the a b term in (a 10
+ b) is 252. Substitute for a and 2c for b to find the coefficient for the sixth term in the original binomial expansion.
5 6
Thus, the binomial coefficient of the a b term in (a + b)
11
is 462. Substitute
p for a and q for b to
find the coefficient for the seventh term in the original binomial expansion. Therefore, the coefficient of the sixth term in the expansion of (
53.
+ 2c)
10
is 32,256
.
, 7th term
SOLUTION:
Therefore, the coefficient of the seventh term in the
The seventh term in the expansion of
expansion of
can be found by evaluating a
b for
54.
, 6th term
SOLUTION:
eSolutions Manual - Powered by Cognero
.
n –r r
n = 11 and r = 7 – 1 or 6.
For
is
n
to have the form (a + b) , let a =
Page 16
The sixth term in the expansion of n –r r
can
Therefore, the coefficient of the seventh term in the
Therefore, the coefficient of the seventh term in the
10-5expansion The Binomial Theorem of is . 54.
Use the Binomial Theorem to expand and simplify each power of a complex number. 55. (i + 2)4
, 6th term
SOLUTION:
4
Apply the Binomial Theorem to expand (a + b) , where a = i and b = 2.
n –r r
b for n = 11 and r = 6
For
SOLUTION:
can
The sixth term in the expansion of be found by evaluating a – 1 or 5.
is −112,266.
expansion of
n
to have the form (a + b) , let a =
and b = −3
. The coefficient of the term
n –r r
56. (i – 3)3
n
containing a b in the expansion of (a + b) is given by nCr. So, to find the binomial coefficient of
SOLUTION: 3
Apply the Binomial Theorem to expand (a + b) , where a = i and b = −3.
6 5
the term containing a b in the expansion of (a + b) 11
, evaluate nCr for n = 11 and r = 5.
57. (1 – 4i)5 5
Apply the Binomial Theorem to expand (a + b) , where a = 1 and b = −4i. 6 5
Thus, the binomial coefficient of the a b term in (a 11
+ b) is 462. Substitute for b for a and −3 to find the coefficient for the sixth term in the original binomial expansion.
58. (2 +
i)
4
SOLUTION: 4
Apply the Binomial Theorem to expand (a + b) , i. where a = 2 and b = Therefore, the coefficient of the seventh term in the expansion of
is −112,266.
Use the Binomial Theorem to expand and simplify each power of a complex number. 55. (i + 2)4
59. SOLUTION:
SOLUTION:
3
4
Apply the Binomial Theorem to expand (a + b) , where a = i and b = 2. eSolutions Manual - Powered by Cognero
Apply the Binomial Theorem to expand (a + b) , where a =
and b = −
. Page 17
where a = 2 and b =
i.
Replacing
−4 yields the expression (−4 +
5
3) . We do not need to apply the Binomial Theorem to simplify this expression.
10-5 The Binomial Theorem
The graph of g(x) is a translation of the graph of f (x). Use the Binomial Theorem to find the polynomial function for g(x) in standard form.
59.
61. f (x) = x4 + 5x
SOLUTION: 3
Apply the Binomial Theorem to expand (a + b) , and b = −
where a =
g(x) = f (x + 3)
.
SOLUTION:
4
Apply the Binomial Theorem to expand (x + 3) and add 5(x + 3).
5
60. (
i + 3)
SOLUTION: Notice that the first term inside the parenthesis can be simplified. Perform this simplification to make applying the Binomial Theorem easier.
62. f (x) = x5 + 1 g(x) = f (x – 4)
Replacing
−4 yields the expression (−4 +
5
3) . We do not need to apply the Binomial Theorem to simplify this expression. The graph of g(x) is a translation of the graph of f (x). Use the Binomial Theorem to find the polynomial function for g(x) in standard form.
61. f (x) = x4 + 5x g(x) = f (x + 3)
SOLUTION: 5
Apply the Binomial Theorem to expand (x – 4) and add 1.
63. f (x) = x6 + 2x3 g(x) = f (x – 1)
eSolutions Manual - Powered by Cognero
Page 18
SOLUTION:
SOLUTION:
Apply the Binomial Theorem to expand and find the
Apply the Binomial Theorem to expand (x – 4) and add 1.
10-5 The Binomial Theorem 63. f (x) = x6 + 2x3 g(x) = f (x – 1)
for h = 0.1, 0.01, 0.001, and 0.0001 and record your results in a table. What do you observe? c. GRAPHICAL Graph the set of resulting 3 functions from part b for f (x) = x on the same coordinate plane. What do you observe? d. ANALYTICAL As h approaches 0, write an expression for the difference quotient when f (x) = n x , where n is a positive integer.
SOLUTION: 3
a. Let f (x) = x .
SOLUTION: Apply the Binomial Theorem to expand and find the 6 3 sum of (x – 1) and 2(x – 1) .
4
Let f (x) = x .
64. f (x) = x7 – 3x4 + 2x
5
g(x) = f (x + 2)
Let f (x) = x .
6
Let f (x) = x .
7
SOLUTION:
Let f (x) = x .
Apply the Binomial Theorem to expand and find the 7 4 sum of (x + 2) , 3(x + 2) , and 2(x + 2). n
Using the pattern, for f (x) = x , the first term will be
65. MULTIPLE REPRESENTATIONS In this problem, you will use the Binomial Theorem to investigate the difference quotient
5
6
7
x , f (x) = x , f (x) = x , f (x) = x , and f (x) = x . Use the pattern to simplify the difference quotient for f (x) n =x . b. TABULAR Evaluate each expression in part a for h = 0.1, 0.01, 0.001, and 0.0001 and record your results in a table. What do you observe? c. GRAPHICAL Graph the set of resulting 3 functions from part b for f (x) = x on the same coordinate plane. What do you observe? d. ANALYTICAL As h approaches 0, write an eSolutions Manual - Powered by Cognero expression for the difference quotient when f (x) = n x , where n is a positive integer.
n –2
h, nC3x
n–
n –1
3 2
h , and so on. The last term will be h . n Therefore, the difference quotient for f (x) = x is nx
a. ANALYTICAL Use the Binomial Theorem to expand and simplify the difference quotient for f (x) = 4
. The next terms will be nC2x
n –1
for power functions.
3
n−1
nx
r r –1
h
+ nC2x
n –2
+ … + h
h + nC3x
n –3 2
h + … + nCrx
n–
n –1
, h ≠ 0.
b. Substitute h = 0.1, h = 0.01, h = 0.001, and h = 0.0001 into each expression found in part a and record the results in a table. For example, when f (x) 3 2 2 = x , the expression became 3x + 3xh + h . Substitute h = 0.1 into the expression.
Repeat this step for each of the expressions and for h = 0.01, h = 0.001, and h = 0.0001. Page 19
terms except the first in each expression get closer to 0. Thus, we can substitute 0 for the terms that have an h.
Substitute h = 0.1 into the expression.
10-5 The Binomial Theorem
Repeat this step for each of the expressions and for h = 0.01, h = 0.001, and h = 0.0001.
66. In the expansion of (ax + b)5 the numerical coefficient of the second term is 400 and the numerical coefficient of the third term is 2000. Find the values of a and b.
SOLUTION: 5
Expand (ax + b) .
4
The coefficient of the second term, 5a b, is equal to 3 2 400 and the coefficient of the third term, 10a b , is equal to 2000. Start with the coefficient of the second term and solve for b.
Substitute b = Sample answer: As h decreases, all the terms except the first in each expression get closer to 0.
3 2
Into 10a b = 2000 and solve for
a.
2
2
c. Graph y = 3x + 0.3x + 0.01, y = 3x + 0.03x + 2 2 0.0001, y = 3x + 0.003x + 0.000001, and y = 3x + 0.0003x + 0.00000001 on the same coordinate plane.
Substitute a = 2 into b =
and solve for b.
Sample answer: The graphs of the functions are almost exactly the same.
d. Find
.
Notice from the table, that as h decreases, all the terms except the first in each expression get closer to 0. Thus, we can substitute 0 for the terms that have an h.
So, a = 2 and b = 5.
67. RESEARCH Although Pascal’s triangle is named
5 eSolutions Manual - Powered Cognero 66. In the expansion ofby (ax + b)
the numerical coefficient of the second term is 400 and the numerical coefficient of the third term is 2000. Find
for Blaise Pascal, other mathematicians applied their knowledge of the triangle hundreds of years before Pascal. Use the Internet or another source to Page 20 research at least one other person who used the properties of the triangle before Pascal was born. Then describe other patterns found in Pascal's
expand [(x + y) + z] . n n n –1 [(x + y) + z] = (x + y) + n(x + y) z + nC2(x + y)
10-5 The Binomial Theorem
n –2 2
z + nC3(x + y)
r r
So, a = 2 and b = 5.
n –3 3
z + . . . + nCr (x + y)
n–
n
z + . . . +z .
67. RESEARCH Although Pascal’s triangle is named
70. PROOF The sums of the coefficients in the first
for Blaise Pascal, other mathematicians applied their knowledge of the triangle hundreds of years before Pascal. Use the Internet or another source to research at least one other person who used the properties of the triangle before Pascal was born. Then describe other patterns found in Pascal's triangle that are not described in this lesson.
five rows of Pascal’s triangle are shown below.
SOLUTION:
of Pascal’s triangle is 2 . (Hint: Write 2 as (1 + 1) . Then use the Binomial Theorem to expand.)
Prove that the sum of the coefficients in the nth row n
Sample answer: Chinese mathematician Chu Shihchieh used the triangle in 1303. If the second number of any row is prime, all other numbers in the row (excluding the 1s) are divisible by the prime. The second diagonal of the triangle is the set of natural numbers. The third diagonal is the set of triangular numbers.
68. ERROR ANALYSIS Jena and Gil are finding the
n
n
SOLUTION: The terms in row n of Pascal’s triangle are nC0, nC1, nC2, .
n
. ., nCn. Show that 2 = nC0 + nC1 + nC2 + . . n
n
. + nCn. Start by writing 2 as (1 + 1) . Then use the Binomial Theorem to expand. n
2
14
6th term of the expansion of (x + y) . Jena says that the coefficient of the term is 3003. Gil thinks that it is 2002. Is either of them correct? Explain your reasoning.
n
= (1 + 1) n 0 n –1 1 n –2 2 = nC0 1 1 + nC1 1 1 + nC2 1 1 0 n
+ . . . + nCn 1 1
= nC0 + nC1 + nC2 + . . . + nCn
SOLUTION: Find the first several terms of the expansion of (x + 14 y) .
71. WRITING IN MATH Describe how to find the
numbers in each row of Pascal’s triangle. Then write a few sentences to describe how the
expansions of (a + b) and (a – b) are different n from the expansion of (a + b) .
n –1
The coefficient of the sixth term is 2002. Gil is correct. Jena evaluated nCr for r = 6 rather than for r = 6 – 1 or 5.
69. CHALLENGE Describe a strategy that uses the n
Binomial Theorem to expand (x + y + z) . Then write and simplify an expansion for the expression.
SOLUTION:
n
expand [(x + y) + z] . n n n –1 [(x + y) + z] = (x + y) + n(x + y) z + nC2(x + n –2 2
r r
SOLUTION: Sample answer: The first and last numbers in each row are 1. Every other number is formed by adding the two numbers just above that number in the n –1
previous row. The expansion of (a + b) has n n terms and the expansion of (a + b) has n + 1 terms. n
Sample answer: Since the Binomial Theorem is used to expand binomials, write the expression x + y + z as a binomial. Treat x + y as a single term and
y)
n
z + nC3(x + y)
n –3 3
z + . . . + nCr (x + y)
n–
n
z + . . . +z .
70. PROOF The sums of the coefficients in the first five rows of Pascal’s triangle are shown below. eSolutions Manual - Powered by Cognero
Also, the coefficients for the expansion of (a + b) come from the nth row of Pascal’s triangle. The n –1 coefficients for the expansion of (a + b) come from the row above the nth row of Pascal’s triangle. The coefficients of the terms of the expansion of (a n
+ b) are all positive, whereas every other n coefficient in the expansion of (a – b) is negative.
72. REASONING Determine whether the following statement is sometimes, always, or never true. Justify your reasoning. If a binomial is raised to the power 5, the two Page 21 middle terms of the expansion have the same coefficients.
from the row above the nth row of Pascal’s triangle. The coefficients of the terms of the expansion of (a n
b) are all positive, whereas every other 10-5+ The Binomial Theorem n coefficient in the expansion of (a – b) is negative.
72. REASONING Determine whether the following statement is sometimes, always, or never true. Justify your reasoning. If a binomial is raised to the power 5, the two middle terms of the expansion have the same coefficients.
SOLUTION:
be equal if a = b. Therefore, the two middle terms of the expansion sometimes have the same coefficients. If the coefficients of each term in the binomial are equal, the two middle terms of the expansion will have the same coefficients.
73. CHALLENGE Explain how you could find a term that does not
in the expansion of
contain the variable v. Then find the term.
SOLUTION: will be
Each term in the expansion of
5
Consider (ax + by) , where a and b are nonzero constants. Expand this binomial. .
of the form
The coefficients of the two middle terms are 3 2 3 2 2 3 2 3 10a b x y and 10a b x y . These terms will only be equal if a = b. Therefore, the two middle terms of the expansion sometimes have the same coefficients. If the coefficients of each term in the binomial are equal, the two middle terms of the expansion will have the same coefficients.
73. CHALLENGE Explain how you could find a term that does not
in the expansion of
contain the variable v. Then find the term.
For v to be completely divided out of the term, the exponent of the v in the numerator must be equal to the exponent of the v in the denominator. So, 7r = 8 − r or r = 1. Since r = 1, this will be the second term of the expansion.
SOLUTION: will be
Each term in the expansion of of the form
.
74. PROOF Use the principle of mathematical induction to prove the Binomial Theorem.
SOLUTION: For v to be completely divided out of the term, the exponent of the v in the numerator must be equal to the exponent of the v in the denominator. So, 7r = 8 − r or r = 1. Since r = 1, this will be the second term of the expansion.
n
nC1
a
n − 1 1
b + nC2 a
n − 2 2
0 n
b + . . . + nCn a b for 1
1 0
any positive integer n. Because (a + b) = 1a b + 0 1 1a b or a + b is a true statement, Pn is true for n = k
k 0
1. Assume that (a + b) = k C0 a b + k C1 a + k C2 a
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n 0
Let Pn be the statement that (a + b) = nC0 a b +
k − 2 2
k − 1 1
b
0 k
b + . . . + k Ck a b is true for a
positive integer k. Show that Pk + 1 must be true. Page 22
n
the principle of mathematical induction, (a + b) = nC0 a 0 n
10-5 The Binomial Theorem
b + nC1 a
n
1 1
b + nC2 a
n−2 2
b + . . . + nCn
a b is true for all positive integers n.
74. PROOF Use the principle of mathematical induction to prove the Binomial Theorem.
Prove that each statement is true for all positive integers n or find a counterexample.
75. 12 + 22 + 32 + … + n 2 =
SOLUTION: n
n 0
Let Pn be the statement that (a + b) = nC0 a b + nC1 a
n 0
n − 1 1
b + nC2 a
n − 2 2
0 n
b + . . . + nCn a b for 1
SOLUTION: Use n = 3.
1 0
any positive integer n. Because (a + b) = 1a b + 0 1 1a b or a + b is a true statement, Pn is true for n = k
k 0
1. Assume that (a + b) = k C0 a b + k C1 a + k C2 a
k − 2 2
k − 1 1
b
0 k
b + . . . + k Ck a b is true for a
positive integer k. Show that Pk + 1 must be true.
Since 14 ≠ 12, the statement is not true.
76. 102n−1 + 1 is divisible by 11. SOLUTION:
2n-1
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn
is true for n = 2, n = 3, and so on. That is, by n
the principle of mathematical induction, (a + b) = n 0
nC0 a b + nC1 a
n
1 1
b + nC2 a
n−2 2
b + . . . + nCn
0 n
a b is true for all positive integers n. Prove that each statement is true for all positive integers n or find a counterexample.
75. 12 + 22 + 32 + … + n 2 = SOLUTION: Use n = 3.
Let Pn be the statement 10 P1 is true because 10
2(1) − 1
+ 1 is divisible by 11.
+ 1 is 11, which is
divisible by 11. Assume Pk is true, where k is a positive integer, and show that Pk + 1 must be true. 2k −1
That is, show that 10
+ 1 = 11r for some integer
2(k + 1) −1
r implies that 10 + 1 is divisible by 11. 2k − 1 10 + 1 = 11r 2k − 1 = 11r – 1 10 2k − 1
100 · 10
= 100(11r – 1)
2k + 1
= 1100r − 100 10 + 1 = 1100r − 99 2(k + 1) −1 10 + 1 = 11(100r − 9) Because r is an integer, 100r − 9 is an integer and 11 2(k +1) − 1 (100r − 9) is divisible by 11. Therefore, 10 is divisible by 11. Because Pn is true for n = 1 and 10
2k + 1
Pk implies Pk + 1, Pn is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 2n−1
10 n.
+ 1 is divisible by 11 for all positive integers
77. GENEAOLGY In the book Roots, author Alex
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Haley traced his family history back many generations. If you could trace your family back for 15 generations, starting with your parents, how many Page 23 ancestors would there be?
SOLUTION:
Pk implies Pk + 1, Pn is true for n = 2, n = 3, and so on. By the principle of mathematical induction,
Then rewrite the vector as a linear combination of the standard unit vectors.
2n−1
1 is divisibleTheorem by 11 for all positive integers 10-510The +Binomial n. Write each equation in standard form. Identify the related conic.
77. GENEAOLGY In the book Roots, author Alex Haley traced his family history back many generations. If you could trace your family back for 15 generations, starting with your parents, how many ancestors would there be?
79. x2+ y 2 − 16x + 10y + 64 = 0 SOLUTION: 2
2
x + y − 16x + 10y + 64 = 0
SOLUTION: Diagram the situation showing the first few generations.
2
Notice that the number of ancestors at each generation forms a geometric sequence, 2, 4, 8, …, 1 2 3 or 2 , 2 , 2 , … . To find the total number of 1
2
3
Because the equation is of the form (x − h) + (y − 2 2 k) = r , the conic is a circle.
80. y 2 +16x − 10y + 57 = 0
15
ancestors, calculate 2 + 2 + 2 + ··· + 2 . Since this is a geometric sequence with r = 2, a 1 = 2, and
SOLUTION:
15
a 15 = 2 , use the formula for the sum of a finite geometric series.
Because only one term is squared, the graph is a parabola.
78. Let
be the vector with initial point D(5, −12), and terminal point E(8, −17). Write as a linear combination of the vectors i and j.
81. x2 + y 2 + 2x + 24y + 141 = 0 SOLUTION:
SOLUTION: First, find the component form. 2
Because the equation is of the form (x − h) + (y − 2 2 k) = r , the conic is a circle. Then rewrite the vector as a linear combination of the standard unit vectors.
Find the partial fraction decomposition of each rational expression.
82. Write each equation in standard form. Identify the related conic. 2
2
79. x + y − 16x + 10y + 64 = 0 SOLUTION: 2
2
x + y − 16x + 10y + 64 = 0
SOLUTION: This rational expression is proper. The denominator has one prime quadratic factor of multiplicity 2. Rewrite the expression as partial fractions. For denominators which are quadratic factors, use Ax + B and Cx + D in the numerators.
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Page 24
2
2
Multiply each side by the LCD, (x – 2) .
the equation is of the form (x − h) 10-5Because The Binomial Theorem 2 2
2
+ (y −
k) = r , the conic is a circle.
Find the partial fraction decomposition of each rational expression.
83. SOLUTION:
82. SOLUTION: This rational expression is proper. The denominator has one prime quadratic factor of multiplicity 2. Rewrite the expression as partial fractions. For denominators which are quadratic factors, use Ax + B and Cx + D in the numerators.
2
2
Multiply each side by the LCD, (x – 2) .
Because the degree of the numerator is greater than or equal to the degree of the denominator, the rational expression is improper. To rewrite the expression, divide the numerator by the denominator.
Therefore, the proper rational expression is
Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the coefficients of the x-terms on the left side of the equation must equal the coefficients of the xterms on the right side.
.
Now we can rewrite the remaining expression as partial fractions with constant numerators, A, and B, and denominators that are the linear factors of the original denominator.
2
Multiply each side by the LCD, x − 8x + 15. Group like terms.
Use any method to solve the new system. Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the coefficients of the x-terms on the left side of the equation must equal the coefficients of the xterms on the right side.
Use any method to solve the new system. Replace A, B, C, and D with 0, 5, 0, and −4 in the partial fraction decomposition.
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83.
Replace A, and B with 2, and 5 in the partial fraction Page 25 decomposition.
10-5 The Binomial Theorem
Replace A, and B with 2, and 5 in the partial fraction decomposition.
Determine whether each matrix is in rowechelon form.
85. The partial fraction decomposition of the improper +
rational expression is x +
SOLUTION:
.
There is a nonzero number below the leading one in the third row. The matrix is not in row-echelon form.
84. SOLUTION: 86.
This rational expression is proper. Because the factor (x − 3) has multiplicity 2, include partial
SOLUTION:
2
fractions with denominators of (x − 3) and (x − 3) .
There is a zero below the leading one in the first row. The matrix is in row-echelon form.
2
Multiply each side by the LCD, (x – 3) .
87. Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the coefficients of the x-terms on the left side of the equation must equal the coefficients of the xterms on the right side.
Use any method to solve the new system.
SOLUTION: There is a zero below the leading one in the first row. The matrix is in row-echelon form. Find the exact value of each trigonometric expression. 88. tan 195°
SOLUTION: Write 195° as the sum or difference of angle measures with targents that you know. Replace A and B with 3 and 9 in the partial fraction decomposition.
Determine whether each matrix is in rowechelon form.
85. SOLUTION: There is a -nonzero below eSolutions Manual Powered number by Cognero
the leading one in the third row. The matrix is not in row-echelon form.
Page 26
SOLUTION: a zero belowTheorem the leading one in the first 10-5There TheisBinomial row. The matrix is in row-echelon form. Find the exact value of each trigonometric expression. 88. tan 195°
SOLUTION: Write 195° as the sum or difference of angle measures with targents that you know.
89. csc SOLUTION: Write
as the sum or difference of angle
measures with csc that you know.
89. csc SOLUTION: Write
as the sum or difference of angle
90. sin
measures with csc that you know.
SOLUTION: Write
as the sum or difference of angle
measures with sines that you know.
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Page 27
SOLUTION:
10-5 The Binomial Theorem
a. Graph elapsed time on the x-axis and balance on the y-axis. Plot the points (0, 30), (5, 41.1), (10, 56.31), (15, 77.16), (20, 105.71), (25, 144.83), (30, 198.43).
90. sin SOLUTION: Write
as the sum or difference of angle
measures with sines that you know. b. Enter the data into the calculator and then select ExpReg.
x
The equation is approximately y = 30(1.065) . c. Use the graph to find the value of y for x = 41.
91. SAVINGS Janet’s father deposited $30 into a bank account for her. They forgot about the money and made no further deposits or withdrawals. The table shows the account balance for several years.
The balance of the account in 41 years will be about $396.70.
92. SAT/ACT In the figure below, ABCD is a parallelogram. What are the coordinates of point C? a. Make a scatter plot of the data. b. Find an exponential function to model the data. c. Use the function to predict the balance of the account in 41 years.
SOLUTION: a. Graph elapsed time on the x-axis and balance on the y-axis. Plot the points (0, 30), (5, 41.1), (10, 56.31), (15, 77.16), (20, 105.71), (25, 144.83), (30, 198.43).
A (d + a, y) B (d – a, b) C (d + x, b) D (d + a, b) E (d + b, a)
SOLUTION: eSolutions Manual - Powered by Cognero
Since ABCD is a parallelogram, must be parallel to . Therefore, the y-coordinate of C must Page be 28 the same as the y-coordinate for B. So, the ycoordinate of C must be b. Also, since ABCD is a
balance of the account in 41 years will be about 10-5The The Binomial Theorem $396.70.
Therefore, the correct answer is F.
92. SAT/ACT In the figure below, ABCD is a parallelogram. What are the coordinates of point C?
94. Mrs. Thomas is giving a four-question multiplechoice quiz. Each question can be answered A, B, C, or D. How many ways could a student answer the questions using each answer A, B, C, or D once? A 20 B 22 C 24 D 26
SOLUTION: A (d + a, y) B (d – a, b) C (d + x, b) D (d + a, b) E (d + b, a)
Use 4! to calculate the number of possible ways of answering the questions. Therefore, the correct answer is C.
95. REVIEW Which expression is equivalent to (2x – 2)
SOLUTION:
4
Since ABCD is a parallelogram, must be parallel to . Therefore, the y-coordinate of C must be the same as the y-coordinate for B. So, the ycoordinate of C must be b. Also, since ABCD is a parallelogram, . Since must be congruent to A is located at the origin and lies on the x-axis, the length of is d. So, the length of must also be d. Since runs parallel to the x-axis and begins at the x-coordinate a, will extend to the x-coordinate d + a. Therefore, the x-coordinate of C must be d + a. Point C is located at (d + a, b). Therefore, the correct answer is D.
? 4 3 2 F 16x + 64x – 96x – 64x + 16 G 16x4 − 32x3 – 192x2 – 64x + 16 4
3
2
H 16x − 64x + 96x – 64x + 16 J 16x4 + 32x3 – 192x2 – 64x + 16
SOLUTION: Use the numbers in the 4th row of Pascal’s triangle 4
to expand (2x–2) .
Therefore, the correct answer is H.
93. REVIEW What is the value of
?
F 495 G 500 H 660 J 710
SOLUTION: Simplify the factorials.
Therefore, the correct answer is F.
94. Mrs. Thomas is giving a four-question multiplechoice quiz. Each question can be answered A, B, C, or D. How- many ways could a student answer the eSolutions Manual Powered by Cognero questions using each answer A, B, C, or D once? A
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