TITLE: Evaluation of ratio of two binomial expression
QUESTION [8 upvotes]: 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

REPLY [4 votes]: We obtain
  \begin{align*}
\color{blue}{A}&\color{blue}{=\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{4k+2}}\\
&=\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{98-4k}\tag{1}\\
&=[z^{98}]\sum_{n=0}^{200}\left(\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{n-4k}\right)z^n\tag{2}\\
&=[z^{98}]\sum_{n=0}^{200}\left(\sum_{{4k+l=n}\atop{k,l\geq 0}}\binom{100}{4k}\binom{100}{l}\right)z^n\\
&=[z^{98}]\frac{1}{4}\left((1+z)^{100}+(1+iz)^{100}\right.\\
&\qquad\qquad\quad\left.+(1-z)^{100}+(1-iz)^{100}\right)(1+z)^{100}\tag{3}\\
&=[z^{98}]\frac{1}{4}\left((1+z)^{200}+\left(1-z^2\right)^{100}\right)\tag{4}\\
&\,\,\color{blue}{=\frac{1}{4}\left[\binom{200}{98}-\binom{100}{49}\right]}\tag{5}
\end{align*}

Comment:

In  (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
In (2) we introduce coefficient of operator and interpret the expression as convolution of the product of two polynomials in $z$.
In (3) we recall the default case $$\sum_{n=0}^{200}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{100}{k}\binom{100}{l}\right)z^n=(1+z)^{100}(1+z)^{100}.$$ We use series multisection with the $4$-th roots of unity to filter all elements which are not a multiple of $4$.
In (4) we skip terms which do not contribute.
In (5) we select the coefficient of $z^{98}$.


We obtain
  \begin{align*}
\color{blue}{B}&\color{blue}{=\sum_{k=1}^{25}\binom{200}{8k-6}}\\
&=\frac{1}{8}\sum_{k=1}^8\left(\omega_{8}^k\right)^6\left(1+\omega_8^k\right)^{200}\tag{6}\\
&=\frac{1}{8}\sum_{k=1}^8\left(\frac{1+i}{\sqrt{2}}\right)^{6k}\left(1+\left(\frac{1+i}{\sqrt{2}}\right)^k\right)^{200}\tag{7}\\
&=\frac{1}{8}\left((-i)(1+\omega_8)^{200}-(1+i)^{200}+i\left(1-\overline{\omega}_8\right)^{200}+(1-1)^{200}\right.\\
&\qquad\qquad\left.(-i)(1-\omega_8)^{200}-(1-i)^{200}+i\left(1+\overline{\omega}_8\right)^{200}+(1+1)^{200}\right)\\
&=\frac{1}{8}\left((1+1)^{200}-\left((1+i)^{200}+(1-i)^{200}\right)\right)\tag{8}\\
&\,\,\color{blue}{=2^{197}-2^{98}}\tag{9}
\end{align*}

Comment:

In (6) we use again multisection of series as we did in (3). This is formula (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
In (7) we note the $8$-th root of unity is $\omega_8=\frac{1+i}{\sqrt{2}}$. We recall the powers of $\omega_8$ modulo $8$: $\{\omega_8,i,-\overline{\omega}_8,-1,-\omega_8,-i,\overline{\omega}_8,1\}$ which are used in the next line.
In (8) we skip terms which do not contribute.


We finally conclude from (5) and (9)
  \begin{align*}
\color{blue}{\frac{A}{B}=\frac{\binom{200}{98}-\binom{100}{49}}{2^{199}-2^{100}}}
\end{align*}