Q52 / 160 · AP EAMCET 2007 · Mathematics
The coefficient of $x^k$ in the expansion of $\frac{1-2 x-x^2}{e^{-x}}$ is
- Correct answerA$\frac{1-k-k^2}{k !}$
- B$\frac{k^2+1}{k !}$
- C$\frac{1-k}{k !}$
- D$\frac{1}{k !}$
Solution
$\begin{aligned} & \text { Now, } \frac{1-2 x-x^2}{e^{-x}} \\ & =\left(1-2 x-x^2\right)\left(e^x\right) \\ & =\left(1-2 x-x^2\right)\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\right. \\ & \left.+\frac{x^k}{k !}+\ldots \infty\right) \\ & =\left(1+x+\frac{x^2}{2 !}+\ldots+\frac{x^k}{k !}+\ldots \infty\right)-2\left(x+x^2+\frac{x^3}{2 !}\right. \\ & \left.+\ldots+\frac{x^k}{(k-1) !}+\frac{x^{k+1}}{k !}+\ldots \infty\right) \\ & -\left(x^2+x^3+\frac{x^4}{2 !}+\ldots .\right. \\ & \left.\frac{x^k}{(k-2) !}+\frac{x^{k+1}}{(k-1) !}+\frac{x^{k+2}}{k !}+\ldots \infty\right) \\ & \end{aligned}$
$\begin{aligned} & \therefore \text { Coefficient of } x^k \text { in }\left(\frac{1-2 x-x^2}{e^{-x}}\right) \\ &=\frac{1}{k !}-\frac{2}{(k-1) !}-\frac{1}{(k-2) !} \\ &=\frac{1}{k !}-\frac{2 k}{k(k-1) !}-\frac{k(k-1)}{k(k-1)(k-2) !} \\ &=\frac{1}{k !}-\frac{2 k}{k !}-\frac{k(k-1)}{k !} \\ &=\frac{1-k-k^2}{k !}\end{aligned}$
Hence, the correct answer is option: 'A'.