Question

n
∑$\frac{2x+1}{k^2(k+1)^2}$ 
k=1

Aduceti la o forma mai simpla.

Answer 1

$\sum\limits_{k=1}^n \dfrac{2k+1}{k^2(k+1)^2} = \sum\limits_{k=1}^n \dfrac{(k+1)^2-k^2}{k^2(k+1)^2} = \\ \\= \sum\limits_{k=1}^{n} \left(\dfrac{(k+1)^2}{k^2(k+1)^2}-\dfrac{k^2}{k^2(k+1)^2}\right) = \\ \\ =\sum\limits_{k=1}^n\left(\dfrac{1}{k^2}-\dfrac{1}{(k+1)^2}\right) = \\ \\ = \sum\limits_{k=1}^n\dfrac{1}{k^2}-\sum\limits_{k=1}^n \dfrac{1}{(k+1)^2} = \\ \\ =\left(\dfrac{1}{1^2}+ \sum\limits_{k=2}^n\dfrac{1}{k^2} \right) - \left(\sum\limits_{k=2}^{n}\dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\right) =$
$= 1 -\dfrac{1}{(n+1)^2} +\sum\limits_{k=2}^n\dfrac{1}{k^2}-\sum\limits_{k=2}^{n}\dfrac{1}{k^2} = \\ \\ = 1-\dfrac{1}{(n+1)^2} = \dfrac{(n+1)^2-1}{(n+1)^2} = \\ \\ = \dfrac{n(n+2)}{(n+1)^2}$