Question

The sum of $n$ terms of the series $\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+.....$ is equal to

• A. $\frac{{{n}^{2}}-2n}{{{(n+1)}^{2}}}$
• B. $\frac{{{n}^{2}}-2}{{{(n+1)}^{2}}}$
• C. $\frac{{{n}^{2}}+2n}{{{(n+1)}^{2}}}$
• D. $\frac{{{n}^{2}}+2}{{{(n+1)}^{2}}}$

Answer 1

Answer: (C)

$\frac{{{n}^{2}}+2n}{{{(n+1)}^{2}}}$

Answer 2

Let ${{S}_{n}}=\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+....+\frac{(2n+1)}{{{n}^{2}}.{{(n+1)}^{2}}}$
$=\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)+\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)$
$+......+\left( \frac{1}{{{n}^{2}}}-\frac{1}{{{(n+1)}^{2}}} \right)$
$=\frac{1}{1}-\frac{1}{{{(n+1)}^{2}}}$
$=\frac{{{n}^{2}}+1+2n-1}{{{(n+1)}^{2}}}$
$=\frac{{{n}^{2}}+2n}{{{(n+1)}^{2}}}$