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} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\
\frac{1276}{1277} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\
& \,\,\,\vdots \\
\frac{2548}{2549} & =\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!

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. 
