Question

The sum to the infinite terms of the series $\frac{5}{3^{2} \cdot 7^{2}}+\frac{9}{7^{2} \cdot 1 1^{2}}+\frac{13}{1 1^{2} \cdot 1 5^{2}}+....$ is

• A. $\frac{1}{8}$
• B. $\frac{1}{36}$
• C. $\frac{1}{54}$
• D. $\frac{1}{72}$

Answer 1

Answer: (D)

$T_{k}=\frac{5 + \left(k - 1\right) 4}{\left(3 + \left(k - 1\right) 4\right)^{2} \left(7 + \left(k - 1\right) 4\right)^{2}}$
$=\frac{4 k + 1}{\left(4 k - 1\right)^{2} \left(4 k + 3\right)^{2}}$
$=\frac{1}{8}\left\{\frac{1}{\left(4 k - 1\right)^{2}} - \frac{1}{\left(4 k + 3\right)^{2}}\right\}$
$S_{n}=T_{1}+T_{2}+T_{3}+....+T_{n}$
$=\frac{1}{8}\left\{\frac{1}{3^{2}} - \frac{1}{7^{2}} + \frac{1}{7^{2}} - \frac{1}{1 1^{2}} + \frac{1}{1 1^{2}} - \frac{1}{1 5^{2}} + . . . . + \frac{1}{\left(4 n - 1\right)^{2}} - \frac{1}{\left(4 n + 3\right)^{2}}\right\}$
$=\frac{1}{8}\left\{\frac{1}{3^{2}} - \frac{1}{\left(4 n + 3\right)^{2}}\right\}$
$S_{\in fty}=\underset{n \rightarrow \in fty}{l i m}S_{n}=\frac{1}{8}\left\{\frac{1}{9} - 0\right\}=\frac{1}{72}$