Show that if X is a Poisson random variable with mean 1, then we have $E(X^n) = E((X + 1)^{n−1})$. Use this result to compute $Var(3X + 1)$ and $E(X^4)$.
Can the value of E(X) be computed by considering the value of n be 1 and then raised the power four be E($X^4$) and Variance would be 9*E($X$)? Is there something wrong with this approach?
Yes. $Var(3X+1)=Var(3X)=9\cdot Var(X)$
But you can calculate $Var(X)=E(X^2)-[E(X)]^2$ with the given formula.
$E(X^n)=E((X + 1)^{n-1})$
Note that the the exponent is $n-1$.
$E(X^2)=E((X+1)^1)=E(X+1)$ and $[E(X)]^2=1$
Second part
$E(X^4)=E((X + 1)^{3})=E(X^3+3X^2+3X+1)=E(X^3)+3E(X^2)+3E(X)+1$
$E(X^3)$ can be calculated by using the given formula as well.