Finding $E(\frac{2^X}{X+1})$, for $X:P(\lambda)$ $X:P(\lambda)$, $P(X=n)=e^{-\lambda} \frac{\lambda^n}{n!}$. Also, $P(X=1)=P(X=3)$. Find $E(\frac{2^X}{X+1})$.
First of all, I don't see how this $P(X=1)=P(X=3)$ can help me..
Next I tried to find the expected value $E(\frac{2^X}{X+1})=\sum_{n=0}\frac{2^n}{n+1}e^{-\lambda} \frac{\lambda^n}{n!}$, but got lost in calculation. Sum goes to infinity.
I also saw in other examples, when computing expected value, that if expression looks like a  derivative, we can replace it, why is it okay to do so? 
 A: From $P(x=1)=P(x=3)$ it follows that $\lambda=\lambda^3/6$, hevce $\lambda=\sqrt{6}$.
Further,
$$E(\frac{2^X}{X+1})=\sum_{n=0}^{+\infty}\frac{2^n}{n+1}e^{-\lambda} \frac{\lambda^n}{n!}=
\sum_{n=0}^{+\infty}\frac{(2\sqrt{6})^n}{(n+1)!}e^{-\sqrt{6}}=\frac{e^{-\sqrt{6}}}{2\sqrt{6}}\sum_{n=0}^{+\infty}\frac{(2\sqrt{6})^{n+1}}{(n+1)!}=$$
$$=\frac{e^{-\sqrt{6}}}{2\sqrt{6}} (e^{2\sqrt{6}}-1).$$
A: following your calculations:
$$E(\frac{2^X}{X+1})=\sum_{n=0}\frac{2^n}{n+1}e^{-\lambda} \frac{\lambda^n}{n!}= e^{-\lambda}\sum_{n=0}^\infty \frac{(2\lambda)^n}{(n+1)!} = \frac{e^{-\lambda}}{2\lambda} \sum_{n=0}^\infty \frac{(2\lambda)^{n+1}}{(n+1)!} =$$ $$= \frac{e^{-\lambda}}{2\lambda} (e^{2\lambda}-1) = \frac{e^\lambda-e^{-\lambda}}{2\lambda}$$
A: $P(X=1)=P(X=3)$ gives the exact value of $\lambda$: $e^{-\lambda} \frac {\lambda^{1}} {1!}=e^{-\lambda} \frac {\lambda^{3}} {3!}$ so $\lambda^{2}=3!$ or $\lambda =\sqrt 6$. 
Now $E\frac {2^{X}} {X+1}= \sum\limits_{k=0}^{\infty} \frac {(2\lambda)^n} {(n+1)!} e^{-\lambda}=e^{-\lambda} \frac 1 {2\lambda} (e^{2\lambda}-1)$/ Put $\lambda =\sqrt 6$ in this. 
