Evaluate the summation

Question

Evaluate the following summation :

\(\sum_{k=0}^{24}\binom{100}{4k}.\binom{100}{4k+2} \)

The answer given is  \(\frac{1}{4} .[\binom{200}{102}- \binom{100}{51} ]\)

Please explain .....

Answer 1

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-kgives
$$S=\sum_{k=0}^{24}\binom{100}{96-4k}\binom{100}{2k+2}.$$
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
$$8S=\binom{200}{98}-2\binom{100}{49}+\binom{200}{98}$$

Rearranging We get 

$$\frac{1}{4} .[\binom{200}{102}-\binom{100}{51}]$$