Binomial random variable, Poisson random variable and probability density function 
Let $\text{X}$ be a binomial random variable with parameters $n$ and $p$. Show that $$E\,\left(\frac{1}{1+X}\right)=\frac{1-(1-p)^{n+1}}{(n+1)p}$$

To be honestly, I have no idea what I should do for this question. It would be great if someone can help me.

Show that if $X$ is a Poisson random variable with parameter $\lambda$, then we have $$E(X^n)=\lambda E((X+1)^{n-1}).$$ Use this result to compute $Var(2X+1)$ and $E(X^3)$.

For this question, is $Var(2X+1) = 4λ$ and $E(X^3) =  λ^3 + 3 λ^2 +  λ$ ? Besides, how to show $E(X^n) = λE((X+1)^{n-1})$ ?

The probability density function of a random variable $X$ is given by $$P(X=i)=K\cdot\frac{2^i}{i!},\,\,\,\,\,i=0,1,2,\cdots,$$ where $K$ is a positive constant. Find $K$ and hence $P(X>2)$.

Same as question 1, I have no idea what I should do for this question. Please advise me.
 A: 1)
Express the expectation as a sum. The following combinatorial identity might be useful:
$$\frac{1}{k+1} \binom{n}{k} = \frac{n!}{(k+1)k!(n-k)!}=\frac{(n+1)!}{(n+1)(k+1)!(n-k)!}=\frac{1}{n+1} \binom{n+1}{k+1}$$
2) Express the expectation as a sum.
$$E[X^n] = \sum_{k=0}^\infty k^n\frac{\lambda^k e^{-\lambda}}{k!}=0+
\lambda\sum_{k=1}^\infty k^{n-1}\frac{\lambda^{k-1} e^{-\lambda}}{(k-1)!}=\lambda\sum_{k=0}^\infty (k+1)^{n-1}\frac{\lambda^{k} e^{-\lambda}}{k!}=E[(X+1)^{n-1}]$$
$$Var[2X+1]=Var[2X] = E[4X^2]-E[2X]^2=4\lambda E[X+1]-4\lambda^2=4\lambda^2+4\lambda-4\lambda^2 = 4\lambda$$
Similarly for the other problem: $E[X^3] = \lambda E[(X+1)^2] = \lambda E[X^2+2X+1] = \dots$
and so on.
3)
For $$P(X=i)=K \frac{2^i}{i!}$$
to be a valid PMF, it must satisfy
$$\sum_{i=0}^\infty P(X=i) = 1$$
Therefore
$$\sum_{i=0}^\infty P(X=i)= \sum_{i=0}^\infty K \frac{2^i}{i!}= K\sum_{i=0}^\infty  \frac{2^i}{i!} = 1$$
What value of $K$ makes the last equality true? (Hint: $\sum_{i=0}^\infty\frac{2^i}{i!}$ is a well-known series).
Once we find $K$, finding $P(X>2)$ is pretty simple, as
$$P(X>2)=\sum_{i=3}^\infty P(X=i) = 1 - P(X=0)-P(X=1)$$
