Does anyone know how to show the following combinatorial equality, $\sum_{i=0}^{n}\left(ni\right)^{2}\binom{2n}{i}=n\cdot4^{n1}$?
By the way, this is not a homework problem, otherwise one would be able to search the answer.
Thanks.
Does anyone know how to show the following combinatorial equality, $\sum_{i=0}^{n}\left(ni\right)^{2}\binom{2n}{i}=n\cdot4^{n1}$? By the way, this is not a homework problem, otherwise one would be able to search the answer. Thanks. 


It's half the sum of the same thing from $0$ to $2n$, which in turn is easily related to the variance of the number of heads in a sequence of $2n$ tosses of a fair coin. 


Try the WilfZeilberger method and its friends. This automatically proves many such (hypergeometric) identities. See the book A = B 


I suggest you take a look on hypergeometric series. 

