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Counting (Binomial)

(CIE 0606/2021/m/22/Q9)
(a) In the expansion of $\left(2 k-\dfrac{x}{k}\right)^{5}$, where $k$ is a constant, the coefficient of $x^{2}$ is 160 . Find the value of $k$. $[3]$

(b) (i) Find, in ascending powers of $x$, the first 3 terms in the expansion of $(1+3 x)^{6}$, simplifying the coefficient of each term.

(ii) When $(1+3 x)^{6}(a+x)^{2}$ is written in ascending powers of $x$, the first three terms are $4+68 x+b x^{2}$, where $a$ and $b$ are constants. Find the value of $a$ and of $b .$ 


*********math solution*************

(b) (i) $\begin{array}[t]{lll}(1+3 x)^{6} &=1+^6 C_{1}(3 x)+^6 C_{2}(3 x)^{2}+\cdots \\&=1+6(3 x)+\frac{6 \times 5}{1 \times 2}\left(9 x^{2}\right)+\cdots \\&=1+18 x+135 x^{2}+\cdots \end{array}$

(ii) $(1+3 x)^{6}(a+x)^{2}$ 

$\begin{array}[t]{rcl}&=&\left(1+18 x+135 x^{2}+\cdots\right)\left(a^{2}+2 a x+x^{2}\right) \\&=&\left(a^{2}+2 a x+x^{2}\right) +\left(18 a^{2} x+36 a x^{2}+18 x^{3}\right) +(135 a x^{2}+\cdots )\\&=& a^{2}+\left(2 a+18 a^{2}\right) x+(1+36 a+135 a) x^{2}+\cdots\end{array}$

$\begin{array}[t]{lll}&\text {Compare with } (1+3x)^6(a+x)^2= 4+68 x+b x^{2}, \\&a^{2}=4, \quad 2 a+18 a^{2}=68, b=1+36 a+135 a \\&a^{2}=4 \Rightarrow a=\pm 2 \qquad\cdots(1)\\&2 a+18 a^{2}=68 \Rightarrow 9 a^{2}+a-34=0 \\&(a+2)(9 a-17)=0 \\&a=-2 \text { or } a=\dfrac{17}{9} \qquad\cdots(2)\\&\text {Hence by (1) and (2), } a=-2 . \\&b=1+36(-2)+135(-2)=469 .\end{array}$
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