Question

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

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

Answer 1

The correct option is A

$\frac{(1-k-k^2)}{k!}$

The explanation for the correct option:

Determining the coefficient of $x^k$

We have the given equation as: $\frac{(1-2x-x^2)}{e^{-x}}$

using expansion to solve the equation.

$$\begin{gathered} \frac{(1–2x–x^2)}{e^{-x}} \\ =(1–2x–x^2)e^x \\ =(1–2x–x^2)\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots ..\right);\left[\because e^x=\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots ..\right)\right] \end{gathered}$$

Coefficient of$x^k=\frac{1}{k!}–\frac{2}{(k-1)!}–\frac{1}{(k-2)!}$

$$\begin{gathered} =\frac{(1-2k-k(k-1)}{k!} \\ =\frac{(1-2k-k^2+k)}{k!} \\ =\frac{(1-k-k^2)}{k!} \end{gathered}$$

Hence option A is correct.