Question

It is given that $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+....=\frac{{\pi }^4}{90}$. Then $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...$is equal to

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

Answer 1

The correct option is A

$\frac{{\pi }^4}{96}$

Explanation for correct option:

Finding the value of given terms:

Let us consider $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}....=S$ .

Therefore we have,

$$\begin{gathered} \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+.....=\frac{{\pi }^4}{90} \\ \text{Let us split the odd and even terms} \\ \Rightarrow \left(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\right)+\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+....\right)=\frac{{\pi }^4}{90} \\ \Rightarrow S+\frac{1}{2^4}\left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+....\right)=\frac{{\pi }^4}{90} \\ \Rightarrow S+\frac{1}{2^4}\times \frac{{\pi }^4}{90}=\frac{{\pi }^4}{90}\left[\text{given}\right] \\ \Rightarrow S=\frac{{\pi }^4}{90}-\frac{{\pi }^4}{90\times 16} \\ \Rightarrow =\frac{{\pi }^4}{90}\left(1-\frac{1}{16}\right) \\ \Rightarrow =\frac{{\pi }^4}{90}\times \frac{15}{16} \\ \Rightarrow S=\frac{{\pi }^4}{96} \end{gathered}$$

Therefore, $S=\frac{{\pi }^4}{96}$.

Hence, option (A) is the correct answer.