# IB Math Analysis & Approaches Questionbank-Topic: SL 1.9-Use of Pascal’s triangle-SL Paper 1

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

evidence of valid binomial expansion with binomial coefficients     (M1)

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

attempt to identify correct term     (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)     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)     (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 $$n$$)     (A1)

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

correct simplification     (A1)

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

$$n = 31$$     A1     N2

OR

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

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

$$n = 31$$     A1     N2

[7 marks]

## Question

The values in the fourth row of Pascal’s triangle are shown in the following table.

Write down the values in the fifth row of Pascal’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     A2     N2

[2 marks]

a.

evidence of binomial expansion with binomial coefficient     (M1)

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

correct substitution into correct term     (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.

$$720{x^3}$$     A1     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 $$720{x^3}$$ is seen previously.

[5 marks]

b.