Evaluation of ratio of two binomial expression

First of all, let us prove that $$A=\frac 14\left(\binom{200}{102}-\binom{100}{51}\right)$$

Proof :

Let us define $C,D,E$ as follows : $$C:= \binom{100}{99}\binom{100}{3}+\binom{100}{95}\binom{100}{7}+\cdots+\binom{100}{3}\binom{100}{99} $$

$$D:= \binom{100}{98} \binom{100}{4}+\binom{100}{94}\binom{100}{8}+\cdots +\binom{100}{2} \binom{100}{100} $$

$$E:= \binom{100}{97}\binom{100}{5}+\binom{100}{93} \binom{100}{9}+\cdots+\binom{100}{5} \binom{100}{97} $$

We get $$A+C+D+E=\sum_{k=0}^{98}\binom{100}{100-k}\binom{100}{2+k}=[x^{102}](1+x)^{200}=\binom{200}{102}\tag1$$

Next, considering $(1+ix)^{100}$ and $(1-ix)^{100}$, we get

$$A-C+D-E=-[x^{102}](1+ix)^{100}(1-ix)^{100}=-[x^{102}](1+x^2)^{100}=-\binom{100}{51}\tag2$$

Also, considering $\left(1+\frac{1+i}{\sqrt 2}x\right)^{100}$ and $\left(1+\frac{1-i}{\sqrt 2}x\right)^{100}$, we get

$$C-E+i(A-D)=[x^{102}]\left(1+\frac{1+i}{\sqrt 2}x\right)^{100}\left(1+\frac{1-i}{\sqrt 2}x\right)^{100}=[x^{102}](1+\sqrt 2x+x^2)^{100}$$ from which $$A-D=0\tag3$$ follows.

It follows from $(1)(2)(3)$ that

$$A=\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{4k+2}=\frac 14\left(\binom{200}{102}-\binom{100}{51}\right)$$

Next, let us prove that $$B=2^{197}-2^{98}$$

Proof :

Let us consider the following sums :

$$\binom n0+\binom n1+\binom n2+\binom n3+\binom n4+\cdots +\binom nn=2^n$$

$$\binom n0-\binom n1+\binom n2-\binom n3+\binom n4-\cdots +(-1)^n\binom nn=0$$

$$-\left(\binom n0+\binom n1i-\binom n2-\binom n3i+\binom n4+\cdots\right)=-(1+i)^n$$

$$-\left(\binom n0-\binom n1i-\binom n2+\binom n3i+\binom n4-\cdots\right)=-(1-i)^n$$

$$-i\left(\binom n0+\binom n1\frac{1+i}{\sqrt 2}+\binom n2i+\frac{-1+i}{\sqrt 2}\binom n3-\binom n4+\cdots\right)=-i\left(1+\frac{1+i}{\sqrt 2}\right)^n$$

$$-i\left(\binom n0-\binom n1\frac{1+i}{\sqrt 2}+\binom n2i-\frac{-1+i}{\sqrt 2}\binom n3-\binom n4+\cdots\right)=-i\left(1-\frac{1+i}{\sqrt 2}\right)^n$$

$$i\left(\binom n0+\frac{1-i}{\sqrt 2}\binom n1-\binom n2i-\frac{1+i}{\sqrt 2}\binom n3-\binom n4+\cdots\right)=i\left(1+\frac{1-i}{\sqrt 2}\right)^n$$

$$i\left(\binom n0-\frac{1-i}{\sqrt 2}\binom n1-\binom n2i+\frac{1+i}{\sqrt 2}\binom n3-\binom n4+\cdots\right)=i\left(1-\frac{1-i}{\sqrt 2}\right)^n$$

Adding these gives

$$8\left(\binom n2+\binom n{10}+\binom n{18}+\cdots \right)$$ $$=2^n-(1+i)^n-(1-i)^n-i\left(1+\frac{1+i}{\sqrt 2}\right)^n-i\left(1-\frac{1+i}{\sqrt 2}\right)^n$$ $$+i\left(1+\frac{1-i}{\sqrt 2}\right)^n+i\left(1-\frac{1-i}{\sqrt 2}\right)^n$$ $$=2^n-2\cdot 2^{n/2}\cos(n\pi/4)-i(2+\sqrt 2)^{n/2}(\cos(n\pi/8)+i\sin(n\pi/8))$$ $$-i(2-\sqrt 2)^{n/2}(\cos(3n\pi/8)-i\sin(3n\pi/8))$$ $$+i(2+\sqrt 2)^{n/2}(\cos(n\pi/8)-i\sin(n\pi/8))$$ $$+i(2-\sqrt 2)^{n/2}(\cos(3n\pi/8)+i\sin(3n\pi/8))$$

$$=2^n-2\cdot 2^{n/2}\cos(n\pi/4)+2(2+\sqrt 2)^{n/2}(\sin(n\pi/8))$$ $$-2(2-\sqrt 2)^{n/2}(\sin(3n\pi/8))$$

from which we have $$\binom n2+\binom n{10}+\binom n{18}+\cdots $$ $$=2^{n-3}-2^{\frac{n-4}{2}}\cos\left(\frac{n\pi}{4}\right)+\frac 14(2+\sqrt 2)^{n/2}\sin\left(\frac{n\pi}{8}\right)-\frac 14(2-\sqrt 2)^{n/2}\sin\left(\frac{3n\pi}{8}\right)$$

For $n=200$, we have $\cos\left(\frac{n\pi}{4}\right)=1$ and $\sin\left(\frac{n\pi}{8}\right)=\sin\left(\frac{3n\pi}{8}\right)=0$, so $$B=\sum_{k=1}^{25}\binom{200}{8k-6}=2^{197}-2^{98}$$

Conclusion :

$$\frac AB=\color{red}{\frac{\binom{200}{102}-\binom{100}{51}}{2^{199}-2^{100}}}$$

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*}