$$\dfrac{\sum_{r=0}^{24} \binom{100}{4r}\binom{100}{4r+2}}{\sum_{r=1}^{25}\binom{200}{8r-6}}$$
The numerator is fine by multiplying 2 sequences we can easily get it but the denominator is taking a long time to be figured out using 8th roots of unity . I am searching for an alternate method to figure out the denominator .Please help
Define $$D:= \sum_{r=1}^{25}\binom{200}{8r-6} = \sum_{r=0}^{24}\binom{200}{8r+2}$$ Then first note that by reindexing and using $\binom{n}{k} = \binom{n}{n-k}$, we have $$\sum_{r=0}^{24}\binom{200}{8r+2} = \sum_{r=0}^{24}\binom{200}{8(24-r)+2} = \sum_{r=0}^{24} \binom{200}{200-(8(24-r)+2)} = \sum_{r=0}^{24}\binom{200}{8r+6}$$
In particular, we note $$2^{199} = \sum_{k=0}^{100}\binom{200}{2r} = 2D+\sum_{r=0}^{50}\binom{200}{4r},$$ so $$\begin{align*}D &= 2^{198}-\frac{1}{2}\sum_{r=0}^{50}\binom{200}{4r} \\&= 2^{198}-\frac{1}{8}(2^{200}+(1+i)^{200}+(1-i)^{200}) \\ &= 2^{197}-2^{98}\end{align*}$$