Solving sum of poisson

Question

I'm trying to solve this summation but I got stuck at the last step. Hope anyone could help me with this algebra.

$\sum\limits_{n=1}^\infty 10000\cdot(n-1)\cdot\frac{1.5^n\cdot e^{-1.5}}{n!}$

Answer 1

Hint:

$$\sum_{n=1}^\infty 10000(n-1)\frac{1.5^ne^{-1.5}}{n!} $$ $$= 10000\sum_{n=1}^\infty \frac{1.5^ne^{-1.5}}{(n-1)!} -10000\sum_{n=1}^\infty \frac{1.5^ne^{-1.5}}{n!} $$ $$ =15000\sum_{n=0}^\infty \frac{1.5^ne^{-1.5}}{n!} -10000\sum_{n=0}^\infty \frac{1.5^ne^{-1.5}}{n!} + 10000 e^{-1.5}$$ and the two sums are obvious from the definition of a Poisson distribution.

Answer 2

Let $a=1.5$. One has $$\sum_{n=1}^{+\infty}\frac{\left(n-1\right)a^n}{n!}=\sum_{n=1}^{+\infty}\frac{na^n}{n!}-\sum_{n=1}^{+\infty}\frac{a^n}{n!}=\sum_{n=1}^{+\infty}\frac{a^n}{\left(n-1\right)!}-\left(\sum_{n=0}^{+\infty}\frac{a^n}{n!}-1\right)$$ $$=a\sum_{n=0}^{+\infty}\frac{a^{n}}{n!}-\left(\sum_{n=0}^{+\infty}\frac{a^n}{n!}-1\right)=1+\left(a-1\right)e^a.$$