Question

The sum of 10 terms of the series $\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots .$ is :

• A. $1$
• B. $\frac{120}{121}$
• C. $\frac{99}{100}$
• D. $\frac{143}{144}$

Answer 1

Answer: (B)

$S=\frac{2^2-1^2}{1^2\times 2^2}+\frac{3^2-2^2}{2^2\times 3^2}+\frac{4^2-3^2}{3^2\times 4^2}+\dots$
$=\left[\frac{1}{1^2}-\frac{1}{2^2}\right]+\left[\frac{1}{2^2}-\frac{1}{3^2}\right]+\left[\frac{1}{3^2}-\frac{1}{4^2}\right]+\dots +\left[\frac{1}{10^2}-\frac{1}{11^2}\right]$
$=1-\frac{1}{121}$
$=\frac{120}{121}$