Let $x$ be a real number and $-2 < x < 2$. When $\frac{x \quad 1}{(x+3)(x-2)}$ is expanded in powers of $x$,…
Let $x$ be a real number and $-2 < x < 2$. When $\frac{x \quad 1}{(x+3)(x-2)}$ is expanded in powers of $x$, then the coefficient of $x^3$ is
- $-\frac{55}{1296}$
- $-\frac{97}{216}$
- $-\frac{13}{216}$
- $-\frac{119}{1800}$
Solution
$\begin{aligned} & \text { } \frac{x+1}{(x+3)(x-2)}=\frac{3}{5} \frac{1}{(x-2)}+\frac{2}{5} \frac{1}{(x+3)} \\ & =-\frac{3}{5 \times 2}\left(1-\frac{x}{2}\right)^{-1}+\frac{2}{5 \times 3}\left(1+\frac{x}{3}\right)^{-1} \\ & \Rightarrow \frac{x+1}{(x+3)(x-2)}=-\frac{3}{10}\left(1+\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3+\right) \\ & +\frac{2}{5}\left(1-\left(\frac{x}{3}\right)+\left(\frac{x}{3}\right)^2-\left(\frac{x}{3}\right)^3+\ldots\right) \\ & \therefore \text { The coefficient of } x^3=-\frac{3}{10}\left(\frac{1}{2}\right)^3-\frac{2}{15} \times \frac{1}{27} \\ & =-\frac{55}{1296}\end{aligned}$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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