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 eaual to ____.

Answer 1

$x=\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots to \infty$
$=\left(\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots to \infty\right)-\left(\frac{1}{2^{4}}+\frac{1}{4^{4}}+\cdots to \infty\right)$
$=\frac{\pi^{4}}{90}-\frac{1}{16}\left(\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\cdots to \infty\right)=\frac{\pi^{4}}{90}-\frac{1}{16}\cdot \frac{\pi^{4}}{90}$
$=\frac{\pi^{4}}{96}.$