To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)

Binomial Expansion Worksheet

Binomial Expansion Worksheet
  • Page 1
 1.  
State the number of terms in the expansion of (k + l)7 and give the first and last terms.
a.
8, k7l, kl7
b.
7, k8, l8
c.
8, k7, l7
d.
7, k7, l7


Solution:

First term: k 7
Last term: l 7
Number of terms: 7 + 1 = 8


Correct answer : (3)
 2.  
Choose the variable parts of the terms of the expansion.(c + d)5
a.
c5, d5
b.
c5, c4d, c3d2, c2d3, cd4, d5
c.
c6, d6
d.
c5d5, c4d4, c3d3, c2d2, cd, d5


Solution:

First term: c5
Last term: d5
Number of terms: 5 + 1 = 6

c5, c4d, c3d2, c2d3, cd4, d5
[Variable parts of the terms of the expansion.]


Correct answer : (2)
 3.  
Write the variable part of the term having a coefficient of 4096. in the expansion (4g - h)6.
a.
x5y
b.
x6
c.
y6
d.
x3y3


Solution:

First term: (4g)6
Last term: (-h)6
Number of terms: 6 + 1 = 7

(4g)6, (4g)5(-h), (4g)4(-h)2, (4g)3(-h)3, (4g)2(-h)4, (4g)(-h)5, (-h)6
[Variable parts of the terms of the expansion.]

So, the variable part of the term having a coefficient of 4096 is g6.


Correct answer : (2)
 4.  
Expand and Simplify: (x12 -y12)5


Solution:

(x12 -y12)5 = (x12 +(-y)12)5
[Write in (a + b)n form.]

(x12)5     (x12)4( - y12)     (x12)³(- y12)²     (x12)²(- y12)³     (x12)(- y12)4     (-y12)5
[Write the variable parts.]

x52           - x2y12               x32y              - xy32                 x12y2              - y52
[Simplify.]

1             5                     10                     10                 5                       1
[Use the Pascal's triangle to find the coefficients.]

To write the expansion, multiply the variable part of each term by its corresponding coefficient.

(x12 -y12)5 = x52 - 5x2y12 + 10x32y - 10xy32 + 5x12y² - y52
[Simplify.]


Correct answer : (0)
 5.  
Expand and Simplify: (5 + i)5
a.
1900 - 5752i
b.
4776i
c.
1900 - 2876i
d.
1900 + 2876i


Solution:

(5 + i)5 = [(5) + (i)]5
[Write in (a + b)n form.]

(5)5     (5)4i     (5)3i2     (5)2i3     (5)i4     i5
[Write the variable parts.]

3125       625i         - 125         - 25i         5         i
[i2 = - 1, i3 = - i, i4 = 1.]

1         5           10             10          5         1
[Use the Pascal's triangle to find the coefficients.]

To write the expansion, multiply the variable part of each term by its corresponding coefficient.

(5 + i)5 = 1(3125) + 5(625i) + 10(- 125) + 10(- 25i) + 5(5) + 1(i)

(5 + i)5 = 3125 + 3125i - 1250 - 250i + 25 + i
[Simplify.]

(5 + i)5 = 1900 + 2876i


Correct answer : (4)
 6.  
Find:
(3 + 2)4 + (3 - 2)4
a.
98
b.
406
c.
98 - 406
d.
98 + 406


Solution:

(3 + 2)4 = [(3) + (2)]4
[Write in (a + b)n form.]

(3)4     (3)3(2)     (3)2(2)2     (3)(2)3     (2)4
[Write the variable parts.]

9                 36                     6                   26              4
[Simplify.]

1                 4                         6                           4                  1
[Use the Pascal's triangle to find the coefficients.]

To write the expansion, multiply the variable part of each term by its corresponding coefficient.

(3 + 2)4 = 1(9) + 4(36) + 6(6) + 4(26) + 1(4)

(3 + 2)4 = 9 + 126 + 36 + 86 + 4
[Simplify.]

(3 + 2)4 = 49 + 206

(3 - 2)4 = [(3) + (-2)]4
[Write in (a + b)n form.]

(3)4     (3)3(- 2)     (3)2(- 2)2     (3) (-2)3     (-2)4
[Write the variable parts.]

9                 - 36                     6                   - 26              4
[Simplify.]

1                 4                           6                           4                  1
[Use the Pascal's triangle to find the coefficients.]

(3 - 2)4 = 1(9) + 4(-36) + 6(6) + 4(-26) + 1(4)

(3 - 2)4 = 9 - 126 + 36 - 86 + 4 = 49 - 206
[Simplify.]

So, (3 + 2)4 + (3 - 2)4 = 49 + 206 + 49 - 206 = 98
[Add both expansions.]


Correct answer : (1)
 7.  
Evaluate:
(1 + 2i)5 + (1 - 2i)5
a.
82
b.
82 - 76i
c.
76i
d.
82 + 76i


Solution:

(1 + 2i)5 = [1 + (2i)]5
[Write in (a + b)n form.]

(1)5     (1)4(2i)     (1)3(2i)2     (1)2(2i)3     (1)(2i)4     (2i)5
[Write the variable parts.]

1         2i           4i2     8i3     16i4     32i5
[Simplify.]

1         2i         - 4     - 8i       16       32i
[i2 = - 1, i3 = - i, i4 = 1, i5 = i.]

1         5           10         10       5         1
[Use the Pascal's triangle to find the coefficients.]

To write the expansion, multiply the variable part of each term by its corresponding coefficient.

(1 + 2i)5 = 1(1) + 5(2i) + 10(-4) + 10(-8i) + 5(16) + 1(32i)

(1 + 2i)5 = 1 + 10i - 40 - 80i + 80 + 32i
[Simplify.]

(1 + 2i)5 = 41 - 38i

(1 - 2i)5 = 1(1) + 5(- 2i) + 10(- 4) + 10(8i) + 5(16) + 1(- 32i)
[Proceed in the same way as above expansion.]

(1 - 2i)5 = 1 - 10i - 40 + 80i + 80 - 32i
[Simplify.]

(1 - 2i)5 = 41 + 38i

So, (1 + 2i)5 + (1 - 2i)5 = 41 - 38i + 41 + 38i
[Add both expansions.]

(1 + 2i)5 + (1 - 2i)5 = 82


Correct answer : (1)
 8.  
A binomial expansion has 12 terms. The variable part of the first term is x11 and the last term is y44. Find the variable parts of the second and third terms.
a.
x10y4, x9y8
b.
x9y8, x11y44
c.
x9y8, x 10y4
d.
x 10y4, x11y44


Solution:

Number of terms = 12 = 11 + 1

First term = x11 = (x)11
Last term = y44 = (y4)11

Therefore, the exponent of the binomial is 11.

Second term is (x)10(y4) = x10y4

Third term is (x)9(y4)2 = x9y8


Correct answer : (1)
 9.  
A binomial expansion has 10 terms. The first term is - x9 and the last term is 19683. Find the third and fourth terms.
a.
9x7, - 27x6
b.
- 9x7, 27x6
c.
3x8, - 9x7
d.
9x7, 27x6


Solution:

Number of terms = 10 = 9 + 1

First term = - x9 = (- x)9
Last term = 19683 = (3)9

Therefore, the exponent of the binomial is 9.

Third term is (- x)7(3)2 = - 9x7

Fourth term is (- x)6(3)3 = 27x6


Correct answer : (2)
 10.  
Find the sum of the coefficients in the expansion of (x²-x-1)23.
a.
2
b.
1
c.
-1


Solution:

Let (x²-x-1)23 = c0 + c1 x + c2 x² + . . . . . . . .

(1-1-1)23 = c0 + c1 + c2 + . . . . . . . . .
[Substitute x = 1 on both the sides.]

Therefore, the sum of the coefficients = (- 1)23 = -1


Correct answer : (4)

*AP and SAT are registered trademarks of the College Board.