Question

The coefficient of $x^{k}$ in the expansion of $\frac{1-2 x-x^{2}}{e^{-x}}$ is

• A. $\frac{1-k-k^{2}}{k !}$
• B. $\frac{k^{2}+1}{k !}$
• C. $\frac{1-k}{k !}$
• D. $\frac{1}{k !}$

Answer 1

Answer: (A)

Now, $\frac{1-2 x-x^{2}}{e^{-x}}$
$=\left(1-2 x-x^{2}\right)\left(e^{x}\right)$
$=\left(1-2 x-x^{2}\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots\right.$
$\left.+\frac{x^{k}}{k !}+\ldots \infty\right)$
$=\left(1+x+\frac{x^{2}}{2 !}+\ldots+\frac{x^{k}}{k !}+. . \infty\right)-2\left(x+x^{2}+\frac{x^{3}}{2 !}\right.$
$\left.+\ldots+\frac{x^{k}}{(k-1) !}+\frac{x^{k+1}}{k !}+\ldots \infty\right)$
$\left.-\frac{x^{k}}{k !}+\ldots \infty\right)$
$-\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}+\ldots .\right.$
$\left.\frac{x^{k}}{(k-2) !}+\frac{x^{k+1}}{(k-1) !}+\frac{x^{k+2}}{k !}+\ldots \infty\right)$
$\therefore $ Coefficient of $x^{k}$ in $\left(\frac{1-2 x-x^{2}}{e^{-x}}\right)$
$=\frac{1}{k !}-\frac{2}{(k-1) !}-\frac{1}{(k-2) !}$
$=\frac{1}{k !}-\frac{2 k}{k(k-1) !}-\frac{k(k-1)}{k(k-1)(k-2) !}$
$=\frac{1}{k !}-\frac{2 k}{k !}-\frac{k(k-1)}{k !}$
$=\frac{1-k-k^{2}}{k !}$