Expectation of 3^x, where X~Binomial(10000,0.5)
Solution 1:
\begin{align} \sum_{i=0}^n3^i(0.5)^{1000}\binom{n}{i}&=(0.5)^n\sum_{i=0}^n3^i\binom{n}{i}\\ &=(0.5)^n\sum_{i=0}^n3^i\cdot 1^{n-i}\cdot \binom{n}{i}\\ &=(0.5)^n (3+1)^n\\ &=2^n \end{align}
where I have used $(a+b)^n = \sum_{i=0}^na^ib^{n-i}\binom{n}{i}$