Given that $ EX^k = (\frac{1}{4}+2^{k-1})$, we have the moment generating function $ M_{X}(t) = \sum _{k=0}^\infty M_{X}^n(0)\frac{t^n}{n!} = \frac{1}{4}+\frac{1}{4}e^t+\frac{1}{2}e^{2t}$
I understand that $M_X^n(0) = EX^n$ and we substitute $(\frac{1}{4}+2^{k-1})$ into the above formula, but not sure how it ends up being equal to : $\frac{1}{4}+\frac{1}{4}e^t+\frac{1}{2}e^{2t}$
It maybe someting simple, but could you please show steps.
Thank you
Your formula can't be right for $k=0$: it would imply $E(X^0)=\frac34$ but of course $E(X^0)=1$. So assume it only holds for $k\ge1$.
As you say, $$M_X(t)=\sum_{k=0}^\infty E(X^k)\frac{t^k}{k!}.$$ Therefore $$M_X(t)=1+\sum_{k=1}^\infty(1/4+2^{k-1})\frac{t^k}{k!} =1+\frac14\sum_{k=1}^\infty\frac{t^k}{k!} +\frac12\sum_{k=1}^\infty\frac{(2t)^k}{k!} =1+\frac{e^t-1}4+\frac{e^{2t}-1}2.$$