Question:
The sum of 10 terms of the series
$\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots . .$ is :
Correct Option: , 2
Solution:
$S=\frac{2^{2}-1^{2}}{1^{2} \times 2^{2}}+\frac{3^{2}-2^{2}}{2^{2} \times 3^{2}}+\frac{4^{2}-3^{2}}{3^{2} \times 4^{2}}+\ldots$
$=\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]+\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]+\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]+\ldots+\left[\frac{1}{10^{2}}-\frac{1}{11^{2}}\right]$
$=1-\frac{1}{121}$
$=\frac{120}{121}$