First slide

Special Series in Sequences and Series

Question

If $\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots to \infty = \frac{π^{2}}{6}, then \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots$ equals

Moderate

A

$π^{2} / 8$
B

$π^{2} / 12$
C

$π^{2} / 3$
D

$π^{2} / 2$

Solution

$\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + \dots$
$= (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \dots) - (\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \dots) \begin{array}{l} = \frac{π^{2}}{6} - \frac{1}{4} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots) \\ = \frac{π^{2}}{6} - \frac{1}{4} (\frac{π^{2}}{6}) \\ = \frac{π^{2}}{8} \end{array}$

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