<s>[INST] 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!
 [/INST] $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550} = \frac{1}{1}-\frac{1}{2} + \frac{1}{3}-\frac{1}{4} + ... + \frac{1}{2549}-\frac{1}{2550}$
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:
$S = 1274 - (H_{2550} - H_{1275}) + \frac{1}{2550} = 1274 - k + \frac{1}{2550}$
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