Problem source

\begin{questionparts}
\item By considering the series expansion of
$(x^2+5x+4){\rm \; e}^x$ show that
\[10{\rm\,  e}=4+\frac{3^2}{1!}+\frac{4^2}{2!}+\frac{5^2}{3!}+\cdots\;.\]
\item Show that
\[5{\rm\,  e}=1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\;.\]
\item Evaluate
\[1+\frac{2^3}{1!}+\frac{3^3}{2!}+\frac{4^3}{3!}+\cdots\;.\]
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
(x^2+5x+4)e^x &= \sum_{k=0}^\infty \frac{1}{k!} x^{k+2}+\sum_{k=0}^\infty \frac{5}{k!} x^{k+1}+\sum_{k=0}^\infty \frac{4}{k!} x^{k} \\
&= \sum_{k=0}^{\infty} \l \frac{1}{k!}+\frac{5}{(k+1)!}+\frac{4}{(k+2)!} \r x^{k+2} + 5x+4+4x \\
&= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{(k+2)(k+1)}{(k+2)!}+\frac{5(k+2)}{(k+2)!}+\frac{4}{(k+2)!} \r x^{k+2}\\
&= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{k^2+3k+2+5k+10+4}{(k+2)!} \r x^{k+2}\\
&= 4 + 9x + \sum_{k=0}^{\infty} \frac{(k+4)^2}{(k+2)!}  x^{k+2}\\
&= 4 + 9x + \sum_{k=2}^{\infty}  \frac{(k+2)^2}{k!}  x^{k}\\
\end{align*}
So when $x = 1$ we have
\[10e = 4 + \frac{3^2}{1!} + \frac{4^2}{2!} + \frac{5^2}{3!} + \cdots \]

\item \begin{align*}
(x^2+3x+1)e^x  &= \sum_{k=0}^\infty \frac{1}{k!}x^{k+2}+\sum_{k=0}^\infty 3\frac{1}{k!}x^{k+1} + \sum_{k=0}^{\infty} \frac{1}{k!} x^k \\
&= 1+3x+\sum_{k=1}^{\infty} \l  \frac1{(k-1)!}+\frac{3}{k!} + \frac{1}{(k+1)!} \r x^{k+1}  \\
&= 1+3x+\sum_{k=1}^{\infty} \frac{(k+1)k + 3(k+1)+1}{(k+1)!}x^{k+1} \\
&= 1+3x+\sum_{k=1}^{\infty} \frac{k^2+4k+4}{(k+1)!}x^{k+1} \\
&= 1+3x+\sum_{k=0}^{\infty} \frac{(k+2)^2}{(k+1)!}x^{k+1} \\
&=1+3x+ \sum_{k=1}^{\infty} \frac{(k+1)^2}{k!}x^k 
\end{align*}
Plugging in $x=1$ we get the desired result.

\item 
\begin{align*} && xe^x &= \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!} \\
x\frac{\d}{\d x} : && x(1+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k+1}}{k!} \\
x\frac{\d}{\d x} : && x(x(1+x)+1+2x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\
&&(x^3+3x^2+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ 
\frac{\d}{\d x} : && e^x(x^3+3x^2+x+3x^2+6x+1) &=\sum_{k=0}^{\infty} \frac{(k+1)^3x^{k}}{k!} \\ 
\Rightarrow && 15e &= 1 + \frac{2^3}{1!} + \frac{3^3}{2!} + \cdots
\end{align*}
\end{questionparts}