I'm supposed to derive a formula for the expected value and variance from the MGF of $X \sim U(a,b)$.
There is a suggested answer going like this:
\begin{align*} M_X(t) &= \frac{1}{t(b-a)}(e^{tb}-e^{ta}) \\ &= \frac{1}{t(b-a)}\left(\sum_{k=0}^n\frac{t^kb^k}{k!}-\sum_{k=0}^n\frac{t^ka^k}{k!}\right) \\ &= ... = 1 + \sum_{k=1}^n\frac{b^{k+1}-a^{k+1}}{(b-a)(k+1)}\frac{t^k}{k!} \end{align*} Which gives
$$E[X^n] = \frac{b^{k+1}-a^{k+1}}{(b-a)(k+1)}$$
However I don't understand the last step, why is $E[X^n] = \frac{b^{k+1}-a^{k+1}}{(b-a)(k+1)}$?
$M_X(t)=Ee^{tX}=E \sum\limits_{k=1}^{\infty} \frac {t^{n}X^{n}} {n!}= \sum\limits_{k=1}^{\infty} \frac {t^{n}EX^{n}} {n!}$. So if you have a series expansion $M_X(t)=\sum a_nt^{n}$ the we get $a_n=\frac {EX^{n}} {n!}$ (or $EX^{n}=a_n n!$) since the coefficients in a power series expansion are unique.