Sec 3 A-Math Binomial theorem worksheet with answers

A free Sec 3 A-Math worksheet on binomial theorem: 8 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Binomial theorem notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Binomial theorem

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 3 A-Math worksheet: Binomial theorem

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

(a)Without using a calculator, find the value of

\,{}^{12}C_{3}

Answer (a)[1]

(b)Use Pascal's triangle to find the coefficient of

x^{3}

in the expansion of

(1 + x)^{4}

Answer (b)[1]

Find the coefficient of

x^{2}

in the expansion of

(3 - x)^{8}

Answer[2]

Given that

\,{}^{n}C_{2} = 153

, find the value of

n

Answer[3]

(a)Find the first three terms in the expansion of

(1 + 5x)^{5}

in ascending powers of

x

Answer (a)[2]

(b)Expand

(3 - x)^{4}

in ascending powers of

x

Answer (b)[3]

(a)In the expansion of

(1 + ax)^{7}

, the coefficient of

x

35

. Find the value of

a

Answer (a)[2]

(b)In the expansion of

(1 + x)^{n}

, the coefficient of

x^2

28

. Find the value of

n

Answer (b)[3]

Section B

(a)Find the term independent of

x

in the expansion of

\left(2x^{2} + \dfrac{1}{x}\right)^{6}

Answer (a)[3]

(b)Find the coefficient of

\dfrac{1}{x^{3}}

in the expansion of

\left(2x^{2} + \dfrac{1}{x}\right)^{6}

Answer (b)[2]

(c)Hence find the term independent of

x

in the expansion of

{(x^{3} + 2)\left(2x^{2} + \dfrac{1}{x}\right)^{6}}

Answer (c)[2]

(a)Find the first three terms in the expansion of

(3 + x)^{6}

in ascending powers of

x

Answer (a)[3]

(b)Given that the coefficient of

x

in the expansion of

{(1 + kx)(3 + x)^{6}}

is zero, find the value of

k

Answer (b)[2]

(c)Using this value of

k

, find the coefficient of

x^{2}

in the expansion of

{(1 + kx)(3 + x)^{6}}

Answer (c)[2]

(a)Without using a calculator, find the value of

\,{}^{8}C_{4}

Answer (a)[2]

(b)Find the first three terms in the expansion of

(1 + 5x)^{7}

in ascending powers of

x

Answer (b)[3]

(c)Find the term independent of

x

in the expansion of

\left(x^{2} - \dfrac{2}{x}\right)^{3}

Answer (c)[4]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

= 220

[1]

\,{}^{12}C_{3} = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1}

= 220

(b)Answer:

\text{Coefficient of } x^{3} = 4

[1]

\def\arraystretch{2}\begin{array}{c} \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 2\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 3\kern0.5em 3\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 4\kern0.5em 6\kern0.5em 4\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \end{array}

\text{Coefficient of } x^{3} = 4

Question 2

Answer:

\text{Coefficient} = 20\,412

[2]

T_{r+1} = {}^{8}C_{r}(3)^{8 - r}(-x)^{r}

r = 2

T_{3} = {}^{8}C_{2}(3)^{6}(-x)^{2}

= 28 \times 729 \times x^{2}

\text{Coefficient} = 20\,412

Question 3

Answer: (

n = -17

is rejected: n can't be negative.)[3]

\dfrac{n(n - 1)}{2} = 153

n(n - 1) = 306

n^2 - n - 306 = 0

(n - 18)(n + 17) = 0

n = 18

(

n = -17

is rejected: n can't be negative.)

Question 4

(a)Answer:

(1 + 5x)^{5} = 1 + 25x + 250x^2 + \ldots

[2]

(1 + 5x)^{5} = {}^{5}C_{0}(1)^{5} + {}^{5}C_{1}(1)^{4}(5x) + {}^{5}C_{2}(1)^{3}(5x)^{2} + \ldots

Term in

x^{0}

{}^{5}C_{0}(1)^{5}(5)^{0} = 1

Term in

x^{1}

{}^{5}C_{1}(1)^{4}(5)^{1} = 25

Term in

x^{2}

{}^{5}C_{2}(1)^{3}(5)^{2} = 250

(1 + 5x)^{5} = 1 + 25x + 250x^2 + \ldots

(b)Answer:

(3 - x)^{4} = 81 - 108x + 54x^2 - 12x^3 + x^4

[3]

(3 - x)^{4} = {}^{4}C_{0}(3)^{4} + {}^{4}C_{1}(3)^{3}(-x) + {}^{4}C_{2}(3)^{2}(-x)^{2} + {}^{4}C_{3}(3)^{1}(-x)^{3} + {}^{4}C_{4}(3)^{0}(-x)^{4}

Term in

x^{0}

{}^{4}C_{0}(3)^{4}(-1)^{0} = 81

Term in

x^{1}

{}^{4}C_{1}(3)^{3}(-1)^{1} = -108

Term in

x^{2}

{}^{4}C_{2}(3)^{2}(-1)^{2} = 54

Term in

x^{3}

{}^{4}C_{3}(3)^{1}(-1)^{3} = -12

Term in

x^{4}

{}^{4}C_{4}(3)^{0}(-1)^{4} = 1

(3 - x)^{4} = 81 - 108x + 54x^2 - 12x^3 + x^4

Question 5

(a)Answer:

a = 5

[2]

T_{r+1} = {}^{7}C_{r}(1)^{7 - r}(ax)^{r}

T_{2} = {}^{7}C_{1}(1)^{6}(ax)^{1}

= 7ax

7a = 35

a = 5

(b)Answer: (

n = -7

is rejected: n can't be negative.)[3]

T_{3} = {}^{n}C_{2}(x)^{2}

= \dfrac{n(n - 1)}{2}x^2

\dfrac{n(n - 1)}{2} = 28

n(n - 1) = 56

n^2 - n - 56 = 0

(n - 8)(n + 7) = 0

n = 8

(

n = -7

is rejected: n can't be negative.)

Question 6

(a)Answer:

60

[3]

T_{r+1} = \dbinom{6}{r}(2x^{2})^{6 - r}\left(\dfrac{1}{x}\right)^{r}

Power of

x

2(6 - r) - r

For the term independent of

x

2(6 - r) - r = 0

r = 4

Term

= \dbinom{6}{4}(2)^{2}(1)^{4}

= 60

(b)Answer:

12

[2]

2(6 - r) - r = -3

r = 5

Coefficient

= \dbinom{6}{5}(2)^{1}(1)^{5}

= 12

(c)Answer:

132

[2]

Term independent of

x

= 2 \times (\text{term independent of } x) + x^{3} \times (\text{term in } \dfrac{1}{x^{3}})

= (2)(60) + 12

= 132

Question 7

(a)Answer:

729 + 1458x + 1215x^{2}

[3]

(3 + x)^{6} = 3^{6} + \dbinom{6}{1}(3)^{5}(x) + \dbinom{6}{2}(3)^{4}(x)^{2} + \ldots

= 729 + 1458x + 1215x^{2} + \ldots

(b)Answer:

k = -2

[2]

(1 + kx)(729 + 1458x + 1215x^{2} + \ldots)

Coefficient of

x

1458 + 729k = 0

k = -2

(c)Answer:

-1701

[2]

Coefficient of

x^{2}

= (1)(1215) + k(1458)

= 1215 + (-2)(1458)

= -1701

Question 8

(a)Answer:

= 70

[2]

\,{}^{8}C_{4} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}

= 70

(b)Answer:

(1 + 5x)^{7} = 1 + 35x + 525x^2 + \ldots

[3]

(1 + 5x)^{7} = {}^{7}C_{0}(1)^{7} + {}^{7}C_{1}(1)^{6}(5x) + {}^{7}C_{2}(1)^{5}(5x)^{2} + \ldots

Term in

x^{0}

{}^{7}C_{0}(1)^{7}(5)^{0} = 1

Term in

x^{1}

{}^{7}C_{1}(1)^{6}(5)^{1} = 35

Term in

x^{2}

{}^{7}C_{2}(1)^{5}(5)^{2} = 525

(1 + 5x)^{7} = 1 + 35x + 525x^2 + \ldots

(c)Answer:

= 12

[4]

T_{r+1} = {}^{3}C_{r}(x^{2})^{3 - r}\left(\dfrac{-2}{x}\right)^{r}

= {}^{3}C_{r}(-2)^{r} x^{6 - 3r}

6 - 3r = 0

r = 2

T_{3} = {}^{3}C_{2}(-2)^{2}

= 3 \times 4

= 12

Practice questions by skill

Sec 3 A-Math Binomial theorem worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Binomial theorem notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)Find the term independent of

x

in the expansion of

\left(2x^{2} + \dfrac{1}{x}\right)^{6}

Answer (a)[3]

(b)Find the coefficient of

\dfrac{1}{x^{3}}

in the expansion of

\left(2x^{2} + \dfrac{1}{x}\right)^{6}

Answer (b)[2]

(c)Hence find the term independent of

x

in the expansion of

{(x^{3} + 2)\left(2x^{2} + \dfrac{1}{x}\right)^{6}}

Answer (c)[2]

Answers and worked solutions

Question 1

(a)Answer:

= 220

[1]

\,{}^{12}C_{3} = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1}

= 220

(b)Answer:

\text{Coefficient of } x^{3} = 4

[1]

\def\arraystretch{2}\begin{array}{c} \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 2\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 3\kern0.5em 3\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \\ \kern1.05em\mathclap{1}\kern1.05em\kern0.5em 4\kern0.5em 6\kern0.5em 4\kern0.5em \kern1.05em\mathclap{1}\kern1.05em \end{array}

\text{Coefficient of } x^{3} = 4

Question 2

Answer:

\text{Coefficient} = 20\,412

[2]

T_{r+1} = {}^{8}C_{r}(3)^{8 - r}(-x)^{r}

r = 2

T_{3} = {}^{8}C_{2}(3)^{6}(-x)^{2}

= 28 \times 729 \times x^{2}

\text{Coefficient} = 20\,412

Question 3

Answer: (

n = -17

is rejected: n can't be negative.)[3]

\dfrac{n(n - 1)}{2} = 153

n(n - 1) = 306

n^2 - n - 306 = 0

(n - 18)(n + 17) = 0

n = 18

(

n = -17

is rejected: n can't be negative.)

Question 4

(a)Answer:

(1 + 5x)^{5} = 1 + 25x + 250x^2 + \ldots

[2]

(1 + 5x)^{5} = {}^{5}C_{0}(1)^{5} + {}^{5}C_{1}(1)^{4}(5x) + {}^{5}C_{2}(1)^{3}(5x)^{2} + \ldots

Term in

x^{0}

{}^{5}C_{0}(1)^{5}(5)^{0} = 1

Term in

x^{1}

{}^{5}C_{1}(1)^{4}(5)^{1} = 25

Term in

x^{2}

{}^{5}C_{2}(1)^{3}(5)^{2} = 250

(1 + 5x)^{5} = 1 + 25x + 250x^2 + \ldots

(b)Answer:

(3 - x)^{4} = 81 - 108x + 54x^2 - 12x^3 + x^4

[3]

(3 - x)^{4} = {}^{4}C_{0}(3)^{4} + {}^{4}C_{1}(3)^{3}(-x) + {}^{4}C_{2}(3)^{2}(-x)^{2} + {}^{4}C_{3}(3)^{1}(-x)^{3} + {}^{4}C_{4}(3)^{0}(-x)^{4}

Term in

x^{0}

{}^{4}C_{0}(3)^{4}(-1)^{0} = 81

Term in

x^{1}

{}^{4}C_{1}(3)^{3}(-1)^{1} = -108

Term in

x^{2}

{}^{4}C_{2}(3)^{2}(-1)^{2} = 54

Term in

x^{3}

{}^{4}C_{3}(3)^{1}(-1)^{3} = -12

Term in

x^{4}

{}^{4}C_{4}(3)^{0}(-1)^{4} = 1

(3 - x)^{4} = 81 - 108x + 54x^2 - 12x^3 + x^4

Question 5

(a)Answer:

a = 5

[2]

T_{r+1} = {}^{7}C_{r}(1)^{7 - r}(ax)^{r}

T_{2} = {}^{7}C_{1}(1)^{6}(ax)^{1}

= 7ax

7a = 35

a = 5

(b)Answer: (

n = -7

is rejected: n can't be negative.)[3]

T_{3} = {}^{n}C_{2}(x)^{2}

= \dfrac{n(n - 1)}{2}x^2

\dfrac{n(n - 1)}{2} = 28

n(n - 1) = 56

n^2 - n - 56 = 0

(n - 8)(n + 7) = 0

n = 8

(

n = -7

is rejected: n can't be negative.)

Question 6

(a)Answer:

60

[3]

T_{r+1} = \dbinom{6}{r}(2x^{2})^{6 - r}\left(\dfrac{1}{x}\right)^{r}

Power of

x

2(6 - r) - r

For the term independent of

x

2(6 - r) - r = 0

r = 4

Term

= \dbinom{6}{4}(2)^{2}(1)^{4}

= 60

(b)Answer:

12

[2]

2(6 - r) - r = -3

r = 5

Coefficient

= \dbinom{6}{5}(2)^{1}(1)^{5}

= 12

(c)Answer:

132

[2]

Term independent of

x

= 2 \times (\text{term independent of } x) + x^{3} \times (\text{term in } \dfrac{1}{x^{3}})

= (2)(60) + 12

= 132

Question 7

(a)Answer:

729 + 1458x + 1215x^{2}

[3]

(3 + x)^{6} = 3^{6} + \dbinom{6}{1}(3)^{5}(x) + \dbinom{6}{2}(3)^{4}(x)^{2} + \ldots

= 729 + 1458x + 1215x^{2} + \ldots

(b)Answer:

k = -2

[2]

(1 + kx)(729 + 1458x + 1215x^{2} + \ldots)

Coefficient of

x

1458 + 729k = 0

k = -2

(c)Answer:

-1701

[2]

Coefficient of

x^{2}

= (1)(1215) + k(1458)

= 1215 + (-2)(1458)

= -1701

Question 8

(a)Answer:

= 70

[2]

\,{}^{8}C_{4} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}

= 70

(b)Answer:

(1 + 5x)^{7} = 1 + 35x + 525x^2 + \ldots

[3]

(1 + 5x)^{7} = {}^{7}C_{0}(1)^{7} + {}^{7}C_{1}(1)^{6}(5x) + {}^{7}C_{2}(1)^{5}(5x)^{2} + \ldots

Term in

x^{0}

{}^{7}C_{0}(1)^{7}(5)^{0} = 1

Term in

x^{1}

{}^{7}C_{1}(1)^{6}(5)^{1} = 35

Term in

x^{2}

{}^{7}C_{2}(1)^{5}(5)^{2} = 525

(1 + 5x)^{7} = 1 + 35x + 525x^2 + \ldots

(c)Answer:

= 12

[4]

T_{r+1} = {}^{3}C_{r}(x^{2})^{3 - r}\left(\dfrac{-2}{x}\right)^{r}

= {}^{3}C_{r}(-2)^{r} x^{6 - 3r}

6 - 3r = 0

r = 2

T_{3} = {}^{3}C_{2}(-2)^{2}

= 3 \times 4

= 12