Question

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+....+\infty =\frac{{\pi }^4}{90}$ then $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+....+\infty =$

• A. $\frac{{\pi }^4}{90}$
• B. $\frac{{\pi }^4}{96}$
• C. $\frac{{\pi }^4}{108}$
• D. $\frac{{\pi }^4}{92}$

Answer 1

The correct option is B $\frac{{\pi }^4}{96}$

Let $x=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+....+\infty =\frac{{\pi }^4}{90}$

Let $y=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+....upto\infty$
So we can write $x=y+\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}.......upto\infty$
$x=y+\frac{1}{16}(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+....upto\infty )$
$x=y+\frac{x}{16}$
$y=\frac{15x}{16}=\frac{15}{16}\times \frac{{\pi }^4}{90}=\frac{{\pi }^4}{96}$