Find the value of $\sum_{i=0}^{24}\binom{200}{4i+2}$ 
Find the value of $$\sum_{i=0}^{24}\binom{200}{4i+2}$$

Actually there was a problem that I was solving and it has a sub part. It was to find the value of $\sum_{r=1}^{25}\binom{200}{8r-6}$
After a lot of rigorous dedication, I transformed the above expression into the original problem. I could not proceed further.
Any help is greatly appreciated.
 A: We will use $i=\sqrt{-1}$. Then define:
$$A=(1+1)^{200}=\sum_{k=0}^{200}{200 \choose k} 1^k$$
$$B=(1-1)^{200}=\sum_{k=0}^{200}{200 \choose k} (-1)^k$$
$$C=(1+i)^{200}=\sum_{k=0}^{200}{200 \choose k} i^k$$
$$D=(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (-i)^k$$
Now take A+B-C-D:
$$2^{200} + 0-(1+i)^{200}-(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (1+(-1)^k-i^k-(-i)^k)$$
Note that $(1+(-1)^k-i^k-(-i)^k)$ is equal to zero unless $k \equiv 2 \pmod 4$, when it is $4$.
$$2^{200} -2 \times 2^{100}=8\sum_{j=0}^{24}{200 \choose {4j+2}} $$
$$\sum_{j=0}^{24}{200 \choose {4j+2}} =2^{197} -  2^{98} $$
A: Consider the polynomial $f(x) = x^2(1+x)^{200}$. Notice that for $i \geq 1$ the coefficient $a_{4i}$ of $x^{4i}$ is $\binom{200}{4(i-1)+2}$. So we would like to get the sum of the coefficients $a_4, a_8, \dots, a_{100}$. Also note that due to the symmetry
$$
\binom{n}{k} = \binom{n}{n-k}
$$
we have
$$
S = \sum_{i = 0}^{24} \binom{200}{4i+2} = \sum_{i=25}^{49} \binom{200}{4i+2}
$$
and so
$$\sum_{i=1}^{25} a_{4i} = \sum_{i=26}^{50} a_{4i}$$
Therefore
$$ S = \frac{1}{2}\sum_{i=1}^{\infty} a_{4i}$$
because bigger coefficients are $0$.
Now let us consider the expression
$$L = \frac{f(1) + f(\omega) + f(\omega^2) + f(\omega^3)}{4},$$
where $\omega = \exp{(\frac{2i\pi}{4})}$ is the fourth root of unity (here $i$ is the imaginary unit). Note that $S = \frac{L}{2}$, based on e.g. this. So we calculate
$$
L = \frac{2^{200} + \omega^2(1+\omega)^{200} + (1+\omega^2)^{200} + \omega^2 (1+\omega^3)^{200}}{4}
$$
Note that $\omega^2 = -1$ and $\omega^2 (1+\omega)^{200} = \omega^2 (1+\omega^3)^{200} = -2^{100}$. So the result is
$$
\frac{L}{2} = \frac{2^{200} - 2\cdot 2^{100}}{8} = 2^{197} - 2^{98}
$$
