Question

The coefficient of $x^n$ in the expansion of $\frac{(1+2x+3x^2)}{e^x}$ is $\frac{(-1{)}^n}{n!}(x{n}^2-(y+z)n+z)$. Find $x+y\times z$

Answer 1

We have $\frac{(a+bx+c{x}^2)}{e^x}=(a+bx+c{x}^2)e^{-x}$

$=(a+bx+c{x}^2)(1+(-x)+\frac{(-x{)}^2}{2!}+...+\frac{(-x{)}^{n-2}}{(n-2)!}+\frac{(-x{)}^{n-1}}{(n-1)!}+\frac{(-x{)}^n}{\left(n\right)!}+...)$

Hence coefficient of $x^n$ in the expansion of $\frac{(a+bx+c{x}^2)}{e^x}$ is

$=\frac{a(-1{)}^n}{n!}+\frac{b(-1{)}^{n-1}}{(n-1)!}+\frac{c(-1{)}^{n-2}}{(n-2)!}$

$=\frac{a(-1{)}^n}{n!}+\frac{b(-1{)}^n}{n(n-1)!}+\frac{c(-1{)}^n(n-1)}{n(n-1)(n-2)!}$

$=\frac{a(-1{)}^n}{n!}+\frac{bn(-1{)}^n}{n!}+\frac{cn(n-1)(-1{)}^n}{n!}$

$=\frac{(-1{)}^n}{n!}(a-bn+c{n}^2-cn)$

$=\frac{(-1{)}^n}{n!}(c{n}^2-(b+c)n+a)$
$\Rightarrow x+y\times z=1+2\times 3=7$