﻿ Binomial Expansion Worksheet | Problems & Solutions

# Binomial Expansion Worksheet

Binomial Expansion Worksheet
• Page 1
1.
State the number of terms in the expansion of ($\mathrm{k + l}$)7 and give the first and last terms.
 a. 8, $k$7$l$, $\mathrm{kl}$7 b. 7, $k$8, $l$8 c. 8, $k$7, $l$7 d. 7, $k$7, $l$7

#### Solution:

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

2.
Choose the variable parts of the terms of the expansion.($\mathrm{c + d}$)5
 a. $c$5, $d$5 b. ${c}^{5}$, ${c}^{4}d$, ${c}^{3}{d}^{2}$, ${c}^{2}{d}^{3}$, c${d}^{4}$, ${d}^{5}$ c. $c$6, $d$6 d. ${c}^{5}{d}^{5}$, ${c}^{4}{d}^{4}$, ${c}^{3}{d}^{3}$, ${c}^{2}{d}^{2}$, cd, ${d}^{5}$

#### 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.]

3.
Write the variable part of the term having a coefficient of 4096. in the expansion (4$\mathrm{g - h}$)6.
 a. $x$5$y$ b. $x$6 c. $y$6 d. ${x}^{3}{y}^{3}$

#### 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.

4.
Expand and Simplify: ()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.]

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

#### 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

6.
Find:
($\sqrt{3}$ + $\sqrt{2}$)4 + ($\sqrt{3}$ - $\sqrt{2}$)4
 a. 98 b. 40$\sqrt{6}$ c. 98 - 40$\sqrt{6}$ d. 98 + 40$\sqrt{6}$

#### 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

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

#### 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

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

8.
A binomial expansion has 12 terms. The variable part of the first term is $x$11 and the last term is $y$44. Find the variable parts of the second and third terms.
 a. $x$10$y$4, $x$9$y$8 b. $x$9$y$8, $x$11$y$44 c. $x$9$y$8, $x$ 10$y$4 d. $x$ 10$y$4, $x$11$y$44

#### 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

9.
A binomial expansion has 10 terms. The first term is - $x$9 and the last term is 19683. Find the third and fourth terms.
 a. 9$x$7, - 27$x$6 b. - 9$x$7, 27$x$6 c. 3$x$8, - 9$x$7 d. 9$x$7, 27$x$6

#### 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

10.
Find the sum of the coefficients in the expansion of ${\left(x²-x-1\right)}^{23}$.
 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