# IB Math Analysis & Approaches Questionbank-Topic: SL 1.9-The binomial theorem -SL Paper 1

## Question

The fifth term in the expansion of the binomial $${(a + b)^n}$$ is given by $$\left( {\begin{array}{*{20}{c}} {10}\\ 4 \end{array}} \right){p^6}{(2q)^4}$$ .

Write down the value of $$n$$.

[1]
a.

Write down a and b, in terms of p and/or q.

[2]
b.

Write down an expression for the sixth term in the expansion.

[3]
c.

## Markscheme

$$n = 10$$Â Â Â Â  A1Â Â Â Â  N1

[1 mark]

a.

$$a = p$$ , $$b = 2q$$Â (or $$a = 2q$$ , $$b = p$$Â )Â Â Â Â  A1A1Â Â Â Â  N1N1

[2 marks]

b.

$$\left( {\begin{array}{*{20}{c}} {10}\\ 5 \end{array}} \right){p^5}{(2q)^5}$$Â Â Â Â  A1A1A1Â Â Â Â  N3

[3 marks]

c.

## Question

Expand $${(2 + x)^4}$$ and simplify your result.

[3]
a.

Hence, find the term in $${x^2}$$ in $${(2 + x)^4}\left( {1 + \frac{1}{{{x^2}}}} \right)$$ .

[3]
b.

## Markscheme

evidence of expandingÂ Â Â Â  M1

e.g. $${2^4} + 4({2^3})x + 6({2^2}){x^2} + 4(2){x^3} + {x^4}$$ , $$(4 + 4x + {x^2})(4 + 4x + {x^2})$$

$${(2 + x)^4} = 16 + 32x + 24{x^2} + 8{x^3} + {x^4}$$Â Â Â Â  A2Â Â Â Â  N2

[3 marks]

a.

finding coefficients 24 and 1Â Â Â Â  (A1)(A1)

term is $$25{x^2}$$Â Â Â Â Â A1Â Â Â Â  N3

[3 marks]

b.

## Question

Given thatÂ $${\left( {1 + \frac{2}{3}x} \right)^n}{(3 + nx)^2} = 9 + 84xÂ + \ldots$$ ,Â find the value of n .

## Markscheme

attempt to expand $${\left( {1 + \frac{2}{3}x} \right)^n}$$Â Â Â  Â (M1)

e.g. Pascal’s triangle, $${\left( {1 + \frac{2}{3}x} \right)^n} = 1 + \frac{2}{3}nx + \ldots$$

correct first two terms of $${\left( {1 + \frac{2}{3}x} \right)^n}$$Â (seen anywhere)Â Â Â Â  (A1)

e.g. $$1 + \frac{2}{3}nx$$

correct first two terms of quadratic (seen anywhere)Â Â Â Â  (A1)

e.g. 9 , $$6nx$$ , $$(9 + 6nx + {n^2}{x^2})$$

correct calculation for the x-termÂ Â Â Â  A2

e.g. $$\frac{2}{3}nx \times 9 + 6nx$$ , $$6n + 6n$$ , $$12n$$

correct equationÂ Â Â Â  A1

e.g. $$6n + 6n = 84$$ , $$12nx = 84x$$

$$n = 7$$Â Â Â Â  A1Â Â Â Â  N1

[7 marks]

## Question

In the expansion of $${(3x + 1)^n}$$, the coefficient of the term in $${x^2}$$ is $$135n$$, where $$n \in {\mathbb{Z}^ + }$$. Find $$n$$.

## Markscheme

Note: Â  Â  Accept sloppy notation (such as missing brackets, or binomial coefficient which includes $${x^2}$$).

evidence of valid binomial expansion with binomial coefficients Â  Â  (M1)

eg$$\;\;\;\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(3x)^r}{(1)^{n – r}},{\text{ }}{(3x)^n} + n{(3x)^{n – 1}} + \left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){(3x)^{n – 2}} +Â \ldots ,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(1)^{n – r}}{(3x)^r}$$

attempt to identify correct term Â  Â  (M1)

eg$$\;\;\;\left( {\begin{array}{*{20}{c}} n \\ {n – 2} \end{array}} \right),{\text{ }}{(3x)^2},{\text{ }}n – r = 2$$

setting correct coefficient or term equal to $$135n$$ (may be seen later) Â  Â  A1

eg$$\;\;\;9\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) = 135n,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ {n – 2} \end{array}} \right){(3x)^2} = 135n,{\text{ }}\frac{{9n(n – 1)}}{2} = 135n{x^2}$$

correct working for binomial coefficient (using $$_n{C_r}$$ formula) Â  Â  (A1)

eg$$\;\;\;\frac{{n(n – 1)(n – 2)(n – 3) \ldots }}{{2 \times 1 \times (n – 2)(n – 3)(n – 4) \ldots }},{\text{ }}\frac{{n(n – 1)}}{2}$$

EITHER

evidence of correct working (with linear equation in $$n$$) Â  Â  (A1)

eg$$\;\;\;\frac{{9(n – 1)}}{2} = 135,{\text{ }}\frac{{9(n – 1)}}{2}{x^2} = 135{x^2}$$

correct simplification Â  Â  (A1)

eg$$\;\;\;n – 1 = \frac{{135 \times 2}}{9},{\text{ }}\frac{{(n – 1)}}{2} = 15$$

$$n = 31$$ Â  Â  A1 Â  Â  N2

OR

evidence of correct working (with quadratic equation in $$n$$) Â  Â  (A1)

eg$$\;\;\;9{n^2} – 279n = 0,{\text{ }}{n^2} – n = 30n,{\text{ (9}}{{\text{n}}^2} – 9n){x^2} = 270n{x^2}$$

evidence of solving Â  Â  (A1)

eg$$\;\;\;9n(n – 31) = 0,{\text{ }}9{n^2} = 279n$$

$$n = 31$$ Â  Â  A1 Â  Â  N2

[7 marks]

## Question

The values in the fourth row of Pascalâ€™s triangle are shown in the following table.

Write down the values in the fifth row of Pascalâ€™s triangle.

[2]
a.

Hence or otherwise, find the term in $${x^3}$$ in the expansion of $${(2x + 3)^5}$$.

[5]
b.

## Markscheme

1, 5, 10, 10, 5, 1 Â  Â  A2 Â  Â  N2

[2 marks]

a.

evidence of binomial expansion with binomial coefficient Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){a^{n – r}}{b^r}$$, selecting correct term,Â $${(2x)^5}{(3)^0} + 5{(2x)^4}{(3)^1} + 10{(2x)^3}{(3)^2} + Â \ldots$$

correct substitution into correct term Â  Â  (A1)(A1)(A1)

eg$$\,\,\,\,\,$$$$10{(2)^3}{(3)^2},{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){(2x)^3}{(3)^2}$$

Note: Award A1 for each factor.

$$720{x^3}$$Â Â  Â  A1 Â  Â  N2

Notes: Do not award any marks if there is clear evidence of adding instead of multiplying.

Do not award final A1 for a final answer of 720, even if $$720{x^3}$$Â is seen previously.

[5 marks]

b.