Binomial Theorem Question 11
Question 11 - 2024 (31 Jan Shift 1)
In the expansion of $(1+x)\left(1-x^{2}\right)\left(1+\frac{3}{x}+\frac{3}{x^{2}}+\frac{1}{x^{3}}\right)^{5}, x \neq 0$, the sum of the coefficient of $x^{3}$ and $x^{-13}$ is equal to
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Answer (118)
Solution
$(1+x)\left(1-x^{2}\right)\left(1+\frac{3}{x}+\frac{3}{x^{2}}+\frac{1}{x^{3}}\right)^{5}$
$=(1+x)\left(1-x^{2}\right)\left(\left(1+\frac{1}{x}\right)^{3}\right)^{5}$
$=\frac{(1+x)^{2}(1-x)(1+x)^{15}}{x^{15}}$
$=\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}$
$=\operatorname{coeff}\left(x^{3}\right)$ in the expansion $\approx \operatorname{coeff}\left(x^{18}\right)$ in
$(1+x)^{17}-x(1+x)^{17}$
$=0-1$
$=-1$
coeff $\left(x^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(x^{2}\right)$ in
$(1+x)^{17}-x(1+x)^{17}$
$=\left(\begin{array}{c}17 \\ 2\end{array}\right)-\left(\begin{array}{c}17 \\ 1\end{array}\right)$
$=17 \times 8-17$
$=17 \times 7$
$=119$
Hence Answer $=119-1=118$
