If 1-14 - 1-24 - 1-34 - - - to - - - 4-90- then 1-14 - 1-34 - 1-54 - - - to - is equals to

The correct option is A π ^4/96 1 / 1 ^ 4 + 1 / 3 ^ 4 + 1 / 5 ^ 4 +...+∞ =( 1 / 1 ^ 4 + 1 / 2 ^ 4 + 1 / 3 ^ 4 +...+∞ nbsp;) ( 1 / 2 ^ ...

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If 1/14 + 1...

Question

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . . . . . . . . +$ to $\infty = \frac{π^{4}}{90}, t h e n \frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . . . . . . . . + t o \infty$ is equals to

\frac{π^{4}}{96}

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\frac{π^{4}}{45}

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\frac{π^{4}}{90}

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none of these

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Solution

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Similar questions

\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . . + \infty = \frac{π^{4}}{90}

\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . . + \infty =

\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots upto \infty = \frac{π^{4}}{90},

\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \dots upto \infty

\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . . . . \infty = \frac{π^{2}}{24}

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . . . \infty

2^{1 / 14} {.4}^{1 / 8} {.8}^{1 / 16} {.16}^{1 / 32} . . . \infty = 2^{x}

x

\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} = . . . \infty

1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + . . .

\infty

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