How to evaluate the following series: $$\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{4k+2}$$
What I have tried : Considering expansion of $\displaystyle (1+x)^n= \binom{n}0+ \binom n1 x + \binom n2 x^2+\cdots$
By this I can get easily the result :
$$\sum_{k=0}^{100}\binom{100}{k}\binom{100}{100-k}=\binom{200}{100}$$
Which seems to be almost similar to $\displaystyle \sum_{k=0}^{24}\binom{100}{4k}\binom{100}{98-4k}$.
I think what I should do next is substitute $i$ and $-i$ and add equations.
What I am not sure is which ones to add. Any suggestion or other approaches? Much Appreciated.
Call the sum $S$. Using the symmetry of Pascal's triangle, $S$ equals $$\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{98-4k}.$$ Putting $j=24-k$ gives $$S=\sum_{k=0}^{24}\binom{100}{96-4k}\binom{100}{2k+2}.$$ Therefore $$2S=\sum_{r=0}^{49}\binom{100}{2r}\binom{100}{98-2r}.$$ This is the $X^{98}$ coefficient of $F(X)^2$ where $$F(X)=\sum_{r=0}^{50}\binom{100}{2r}X^{2r}=\frac{(1+X)^{100}+(1-X)^{100}}{2}.$$ Then $$F(X)^2=\frac{(1+X)^{200}+2(1-X^2)^{100}+(1-X)^{200}}4.$$ I get $$8S=\binom{200}{98}-2\binom{100}{49}+\binom{200}{98}$$ etc.