If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $\vert a-b\vert$ (where $a$ & $b$ are relatively prime) The following question is taken from the practice set of JEE exam.

If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $|a-b|$  (where $a$ & $b$ are relatively prime).

I tried something but I don't think it's the right way to go:
$$\dfrac1{200}\sum_{n=1}^{399}\dfrac{1}{5^{n-200}+1}\\=\dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{1}{5^{n-200}+1}+\dfrac12+\sum_{n=201}^{399}\dfrac{1}{5^{n-200}+1}\right)\\=\dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{1}{5^{n-200}+1}+\dfrac12+\sum_{n=1}^{199}\dfrac{1}{5^{n}+1}\right)$$
 A: You have,
$\dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{1}{5^{n-200}+1}+\dfrac12+\sum_{n=1}^{199}\dfrac{1}{5^{n}+1}\right)$
$ = \dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{5^{200-n}}{5^{200-n}+1}+\dfrac12+\sum_{n=1}^{199}\dfrac{1}{5^{n}+1}\right)$
$ = \dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{5^n}{5^n+1}+\dfrac12+\sum_{n=1}^{199}\dfrac{1}{5^{n}+1}\right)$
$ = \dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{5^n + 1}{5^n+1}+\dfrac12\right) = \cfrac{399}{400}$
A: a very common trick among JEE summation questions is: pair up the first and last terms
$$\sum_{n=1}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+\sum_{n=1, n\not=200}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+ \sum_{n=1}^{199} \frac{5^{200}}{5^{200}+5^n}+\frac{5^{200}}{5^{200}+5^{400-n}}$$ and now an amazing thing happens $\frac{5^{200}}{5^{200}+5^n}+\frac{5^{200}}{5^{200}+5^{400-n}}=1$. So our sum is $\frac{1}{2}+199 \cdot 1=\frac{399}{2}$ $$\implies \frac{a}{b}=\frac{1}{200} \cdot \frac{399}{2}=\frac{399}{400} \implies |a-b|=1$$
A: Whatever the base (A), the relation is always valid:
$\frac{2}{m}\sum_{n=1}^{m-1}\dfrac{A^{m/2}}{A^n+A^{m/2}}=\dfrac {m-1}{m}$
