Question

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

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

Answer 1

The correct option is D none of these

Given that, $1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+......\infty =\frac{{\pi }^4}{90}$
$\Rightarrow (\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\infty )+(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\frac{1}{8^4}+...\infty )=\frac{{\pi }^4}{90}$
$\Rightarrow (\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\infty )+\frac{1}{2^4}(1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+......\infty )=\frac{{\pi }^4}{90}$
$\Rightarrow (\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\infty )+\frac{{\pi }^4}{90\times 2^4}=\frac{{\pi }^4}{90}$
Therefore, $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+...\infty =\frac{{\pi }^4}{90}(1-\frac{1}{2^4})=\frac{{\pi }^4}{96}$