Question

Find the coefficient of $x^n$ in the expansion of $\left(\frac{1+ax+x^2}{e^x}\right)$

• A. $\frac{(-1{)}^n}{n!}1-ann(n-1)$
• B. $\frac{(-1{)}^n}{n!}[1-(n+a)(n-1)]$
• C. $\frac{(-1{)}^n}{n!}1-ann(n+1)$
• D. None of these

Answer 1

Answer: (B)

$\frac{(-1{)}^n}{n!}[1-(n+a)(n-1)]$

Let $z=\left(\frac{1+ax+x^2}{e^x}\right)=(1+ax+x^2)e^{-x}$

$=(1+ax+x^2)(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+..........)$

$\therefore$ Coefficient of $x^n$ in $z$ is $=1.(-1{)}^n\frac{1}{n!}+a.(-1{)}^{n-1}\frac{1}{(n-1)!}+1.(-1{)}^{n-2}\frac{1}{(n-2)!}$

$=\frac{(-1{)}^n}{n!}(1-a(n-1)+n(n-1\left)\right)=\frac{(-1{)}^n}{n!}[1-(n-1\left)\right(n+a\left)\right]$