Question

If $1 +\frac{1}{2^{2}} + \frac{1}{3^{2}} +\frac{1}{4^{2}} + ......\infty = \frac{\pi^{2}}{6}$then the value of $1 -\frac{1}{2^{2}} +\frac{1}{3^{2}} - \frac{1}{4^{2}} + \frac{1}{5^{2}} - \frac{1}{6^{2}} + ......\infty$ equals to

• A. $\frac{\pi^{2}}{24}$
• B. $\frac{\pi^{2}}{48}$
• C. $\frac{\pi^{2}}{12}$
• D. $\frac{5\pi^{2}}{12}$

Answer 1

Answer: (C)

Given,
$1 +\frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + ......\infty = \frac{\pi^{2}}{6} =A (say) ...\left(i\right)$
$\therefore 1 -\frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + .......\infty$
$= \left(1 +\frac{1}{2^{2}} + \frac{1}{3^{2}} +\frac{1}{4^{2}} + .....\infty\right) -2 \left(\frac{1}{2^{2}} + \frac{1}{4^{2}} +\frac{1}{6^{2}} + ......\infty\right)$
$= \left(1 +\frac{1}{2^{2}} + \frac{1}{3^{2}} + ....\infty\right) - \frac{2}{2^{2}} \left(1 +\frac{1}{2^{2}} + \frac{1}{3^{2}} + .....\infty\right)$
$ =A - \frac{1}{2}A = \frac{A}{2} = \frac{\pi^{2}}{12}$ (Using (i))