If $\displaystyle A = \sum_{k=0}^{24}\binom{100}{4k}.\binom{100}{4k+2}$ and $\displaystyle B = \sum_{k=1}^{25}\binom{200}{8k-6}.$ Then $\displaystyle \frac{A}{B}$
$\bf{My\; Try::}$ For evaluation of $$A= \sum_{k=0}^{24}\binom{100}{4k}.\binom{100}{4k+2}= \sum^{24}_{k=0}\binom{100}{100-4k}\cdot \binom{100}{4k+2}$$
$$ = \binom{100}{100}\cdot \binom{100}{2}+\binom{100}{96}\cdot \binom{100}{6}+\cdots \cdots+\binom{100}{4}\cdot \binom{100}{98} = \binom{200}{102}$$
Using $$(1+x)^{100} = \binom{100}{0}+\binom{100}{1}x+\binom{100}{2}x^2+\cdots +\binom{100}{100}x^{100}$$
and $$(x+1)^{100} = \binom{100}{0}x^{100}+\binom{100}{1}x^{99}+\binom{100}{2}x^2+\cdots +\binom{100}{100}$$
Now finding Coefficients of $x^{102}$ in $\displaystyle (1+x)^{100}\cdot (x+1)^{100} = \binom{200}{102}$
Now how can i calculate $B,$ Help Required, Thanks