Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.

Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.

I've tried to write the terms of $S$ in the following way:

\begin{align}
\frac{1275}{1276} &amp; =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\
\frac{1276}{1277} &amp; =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\
&amp; \,\,\,\vdots \\
\frac{2548}{2549} &amp; =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2548\times2549}
\end{align}

But it's just not enough, because in $k$ we don't have $\frac{1}{2\times3}$, $\frac{1}{4\times5}$, etc.

Can you please help me solve the problem? Thanks!
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 
$$\forall n\geq 2,S_n=n-1+\frac1{2n}-k_n.$$
Note that your sum is $$\sum_{n=1}^{1275}{\frac{1}{2n(2n-1)}}$$
We can rewrite this using partial fractions as:
$$\sum_{n=1}^{1275}{\bigg[\frac{1}{2n-1}-\frac{1}{2n}\bigg]}$$
and use telescoping from there.