# Given $\large\frac{1}{1^4}+\large\frac{1}{2^4}+\large\frac{1}{3^4}$+.....$\infty$ = $\large\frac{{\pi}^{4}}{90}$, then the value of $\large\frac{1}{1^4}+\large\frac{1}{3^4}+\large\frac{1}{5^4}$+....$\infty$ is

$(a)\;\large\frac{{\pi}^{4}}{45}\qquad(b)\;\large\frac{{\pi}^{4}}{180}\qquad(c)\;\large\frac{{\pi}^{4}}{96}\qquad(d)\;\large\frac{89{\pi}^{4}}{90}$

Answer : (c) $\;\large\frac{{\pi}^4}{96}$
Explanation : $\;\large\frac{1}{1^4}+\large\frac{1}{2^4}+\large\frac{1}{3^4}+.....=\large\frac{{\pi}^{4}}{90}$
$\;(\large\frac{1}{1^4}+\large\frac{1}{2^4}+\large\frac{1}{3^4}+...)+(\large\frac{1}{2^4}+\large\frac{1}{4^4}+\large\frac{1}{6^4})+....)=\large\frac{{\pi}^{4}}{90}$
$\;(\large\frac{1}{1^4}+\large\frac{1}{3^4}+\large\frac{1}{5^4}+...)+\large\frac{1}{2^4}(\large\frac{1}{1^4}+\large\frac{1}{2^4}+\large\frac{1}{3^4})+....)=\large\frac{{\pi}^{4}}{90}$
$\;(\large\frac{1}{1^4}+\large\frac{1}{3^4}+\large\frac{1}{5^4}+...)+\large\frac{1}{2^4}(\large\frac{{\pi}^{4}}{90})=\large\frac{{\pi}^{4}}{90}$
$\;\large\frac{1}{1^4}+\large\frac{1}{3^4}+\large\frac{1}{5^4}+.....=\large\frac{{\pi}^{4}}{90}\;(1-\large\frac{1}{16})$
$=\large\frac{{\pi}^{4}}{90}\;(1-\large\frac{1}{16})$
$=\large\frac{{\pi}^{4}}{90}\;(\large\frac{15}{16})$
$=\large\frac{{\pi}^{4}}{96}$