First slide

Special Series in Sequences and Series

Question

$If it is given that \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots \infty is \frac{π^{4}}{90} then the value of \frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \dots \infty$

Moderate

A

$\frac{π^{4}}{96}$
B

$\frac{π^{4}}{45}$
C

$\frac{{89π}^{4}}{90}$
D

$\frac{π^{6}}{96}$

Solution

$let x = \frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + ... \infty$
$\begin{array}{l} g i v e n e q u a t i o n i s = \frac{π^{4}}{90} = (\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \dots) + \frac{1}{2^{4}} (\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots) \\ \frac{π^{4}}{90} = x + \frac{1}{16} \cdot \frac{π^{4}}{90} \\ x = \frac{π^{4}}{90} (1 - \frac{1}{16}) \\ = \frac{π^{4}}{90} \times \frac{15}{16} = \frac{π^{4}}{96} \end{array}$

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