Question

If $1/1^4+1/2^4+1/3^4+1/4^4+...\infty ={\pi }^4/90,$ then the value of $\mathbf{1}\mathbf{/}{\mathbf{1}}^{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{/}{\mathbf{3}}^{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{/}{\mathbf{5}}^{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{\infty }$ is

• A. ${\pi }^4/96$
• B. ${\pi }^4/45$
• C. $89{\pi }^4/90$
• D. None of these

Answer 1

The correct option is A

${\pi }^4/96$

Finding the value of :

Given,

$$\begin{gathered} \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+...\infty =\frac{{\pi }^4}{90} \\ \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)+\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+...\right)=\frac{{\pi }^4}{90} \\ \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)+\left(\frac{1}{2^4}\right)\left(1+\frac{1}{2^4}+\frac{1}{3^4}+...\right)=\frac{{\pi }^4}{90} \\ \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)+\left(\frac{1}{2^4}\right)\times \frac{{\pi }^4}{90}=\frac{{\pi }^4}{90} \\ \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)=\frac{{\pi }^4}{90}-\left(\frac{1}{2^4}\right)\times \frac{{\pi }^4}{90} \\ \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)=\frac{{\pi }^4}{90}\left(1-\frac{1}{2^4}\right) \\ =\frac{{\pi }^4}{90}\left(\frac{16-1}{16}\right) \end{gathered}$$

$$\begin{gathered} =\frac{{\pi }^4}{90}\left(\frac{15}{16}\right) \\ =\frac{{\pi }^4}{6}\left(\frac{1}{16}\right) \\ \mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{4}}}{\mathbf{96}} \end{gathered}$$

Hence, option (A) is correct.