Calculate $$\sum_{n=1}^\infty 10,000(n-1)\frac{(3/2)^ne^{-3/2}}{n!}$$
The answer my solution guide gives is:
$$\sum_{n=1}^\infty 10,000(n-1)\frac{(3/2)^ne^{-3/2}}{n!}$$
$$ =\sum_{n=0}^\infty 10,000(n-1)\frac{(3/2)^ne^{-3/2}}{n!} - 10,000(-1)e^{-3/2}$$
$$=10,000(1.5-1)+10,000e^{-3/2}$$
$$=7,231$$
My attempt to get this answer:
$$\sum_{n=1}^\infty 10,000(n-1)\frac{(3/2)^ne^{-3/2}}{n!}$$
$$ =\sum_{n=0}^\infty 10,000(n-1)\frac{(3/2)^ne^{-3/2}}{n!} - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \sum_{n=0}^\infty \frac{(n-1)(3/2)^n}{n!} - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \left(\sum_{n=0}^\infty \frac{n(3/2)^n}{n!}-\sum_{n=0}^\infty \frac{(3/2)^n}{n!}\right) - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \left(\sum_{n=0}^\infty \frac{(2/3)(3/2)^{n-1}}{(n-1)!}-\sum_{n=0}^\infty \frac{(3/2)^n}{n!}\right) - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \left(\frac{2}{3}\sum_{n=0}^\infty \frac{(3/2)^{n-1}}{(n-1)!}-\sum_{n=0}^\infty \frac{(3/2)^n}{n!}\right) - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \left(\frac{2}{3}e^{3/2}-e^{3/2}\right) - 10,000(-1)e^{-3/2}$$
$$= 10,000e^{-3/2} \left(\frac{2}{3}e^{3/2}-e^{3/2}\right) + 10,000e^{-3/2}$$
$$=-1102$$
Let $c=10000\times\exp(-3/2)$. Then the sum is, ignoring the constant factor $c$, $$ \sum_{n=1}^{\infty}\frac{(n-1)(3/2)^{n}}{n!}=\sum_{n=1}^{\infty}\frac{n(3/2)^{n}}{n!}-\sum_{n=1}^{\infty}\frac{(3/2)^{n}}{n!}. $$ Treating each sum separately (by Taylor expansions), $$ \frac{3}{2}e^{3/2}=\frac{3}{2}\sum_{n=0}^{\infty}\frac{(3/2)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(3/2)^{n+1}}{n!}=\sum_{n=1}^{\infty}\frac{(3/2)^{n}}{(n-1)!} $$ and $$ e^{3/2}-1=\sum_{n=0}^{\infty}\frac{(3/2)^{n}}{n!}-1=\sum_{n=1}^{\infty}\frac{(3/2)^{n}}{n!}. $$ You should be able to do the rest to get $\approx 7231.30160148430$.