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Find value of given sum

hxthanh

New member
Sep 20, 2013
16
Put $1\le n\in\mathbb Z$
Find the Sum:
$S_n=\displaystyle \sum_{k=1}^n\dfrac{2k+1-n}{(k+1)^2(n-k)^2+1}$
 

hxthanh

New member
Sep 20, 2013
16
Re: Find the Sum

My solution
Denote $j=n-k-1$ then $k=1 \to j=n-2 \quad ;k=n \to j=-1$.
We get:
\begin{array}{rcl}S_n &=& \sum\limits_{k = 1}^n {\frac{{2k + 1 -n}}{{{{\left( {n - k} \right)}^2}{{\left( {k + 1} \right)}^2} + 1}}} = \sum\limits_{j = - 1}^{n - 2} {\frac{{n - 1 - 2j}}{{{{\left( {j + 1} \right)}^2}{{\left( {n - j} \right)}^2} + 1}}} \\\Rightarrow 2S_n &=& \sum\limits_{k = 1}^n {\frac{{2k + 1 - n}}{{{{\left( {n - k} \right)}^2}{{\left( {k + 1} \right)}^2} + 1}}} + \sum\limits_{k = - 1}^{n - 2} {\frac{{n - 1 - 2k}}{{{{\left( {k + 1} \right)}^2}{{\left( {n - k} \right)}^2} + 1}}} \\&=& n + 1 + \frac{{n - 1}}{{{n^2} + 1}} + \sum\limits_{k = 1}^{n - 2} {\frac{{2k + 1 - n}}{{{{\left( {n - k} \right)}^2}{{\left( {k + 1} \right)}^2} + 1}}} \\&+& \sum\limits_{k = 1}^{n - 2} {\frac{{n - 1 - 2k}}{{{{\left( {n - k} \right)}^2}{{\left( {k + 1} \right)}^2} + 1}}} + \left( { n + 1} \right) + \frac{{n - 1}}{{{n^2} + 1}}\\\Rightarrow S_n &=& n +1 + \frac{{n - 1}}{{{n^2} + 1}}\\&=& \boxed{\displaystyle \frac{{ n\left( {{n^2} + n + 2} \right)}}{{{n^2} + 1}}}\end{array}