Find the value of $S$ in term of $k$ (telescoping sums)

Your sums are $k_{1275}$ and $S_{1275}$ where $$\cases{k_n=\sum_{i=1}^n\frac1{(2i-1)(2i)}\\S_n=\sum_{i=n}^{2n-2}\frac{i}{i+1}}$$

You can prove using induction that $$\forall n\geq 2,S_n=n-1+\frac1{2n}-k_n.$$

$k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550} = \frac{1}{1}-\frac{1}{2} + \frac{1}{3}-\frac{1}{4} + ... + \frac{1}{2549}-\frac{1}{2550}$

Collecting even and odd and with some manipulation, we get

$k = H_{2550} - H_{1275}$ where $H_n$ = harmonic number

$S = 1 - \frac{1}{1276} + 1 - \frac{1}{1277} + ... + 1 - \frac{1}{2549}$

Collecting the ones and with some manipulation we get:

$S = 1274 - (H_{2550} - H_{1275}) + \frac{1}{2550} = 1274 - k + \frac{1}{2550}$