Sums of different infinite series equal the same thing?

Question

Can anyone tell me how $$\sum_{n=0}^\infty \left(\frac{2^{n-2}(n^2-3n+2)}{n!}\right)$$ and $$\sum_{n=0}^\infty \frac{2^{n}}{n!}$$ both equal $$\frac{e^2}{2}$$.

Thank you!

Answer 1

I think your second expression should be $\sum_{n=0}^\infty \frac{2^{n-1}}{n!}$.

Hint: rewrite $\quad n^3-3n+2\quad$ as $\quad n(n-1) -2n + 2$.