$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}} + \frac{7}{3^{2}4^{2}} + \ldots +$ upto $n$ terms is eq

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$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}} + \frac{7}{3^{2}4^{2}} + \ldots +$ upto $n$ terms is equal to

Accepted Answer

Sol. [Key Concept: $\frac{2k + 1}{k^{2(}k{+ 1)}^{2}} = \frac{1}{k^{2}} - \frac{1}{(k{+ 1)}^{2}}$ — telescopes to $1 - \frac{1}{(n{+ 1)}^{2}}$.]$$\sum_{r = 1}^{n}\mspace{2mu}\frac{2r + 1}{r^{2}(r + 1)^{2}} = \sum_{r = 1}^{n}\mspace{2mu}\frac{(r + 1)^{2} - r^{2}}{r^{2}(r + 1)^{2}} = \sum_{r = 1}^{n}\mspace{2mu}\left( \frac{1}{r^{2}} - \frac{1}{(r + 1)^{2}} \right) = 1 - \frac{1}{(n + 1)^{2}}$$

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