If 1-14-1-24-1-34- upto -4-90- then the value of 1-14-1-34-1-54- upto - is

The correct option is B π^49611^4 + 13^4 + 15^4 + 17^4 + nbsp; … = ( 11^4 + 12^4 + 13^4 + 14^4 + 15^4 + nbsp; … ) ( 12^4+14^4+16^4+ … ) nbsp; =π^490 12^ ...

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Byju's Answer

Standard XII

Mathematics

Infinite GP

If 1/14+1/24+...

Question

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots upto \infty = \frac{π^{4}}{90},$ then the value of $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \dots upto \infty$ is

\frac{π^{4}}{45}

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\frac{π^{4}}{96}

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\frac{π^{4}}{124}

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\frac{π^{4}}{196}

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Solution

The correct option is B $\frac{π^{4}}{96}$
$\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \frac{1}{7^{4}} + \dots$
$= (\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \frac{1}{4^{4}} + \frac{1}{5^{4}} + \dots) - (\frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + \dots)$
$= \frac{π^{4}}{90} - \frac{1}{2^{4}} (\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots)$
$= \frac{π^{4}}{90} - \frac{1}{16} (\frac{π^{4}}{90})$
$= \frac{π^{4}}{96}$

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Similar questions

Q. If

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

then

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

Q. If

\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . . . . . . . . +

\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

Q. It is given that

\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .

\infty = \frac{π^{4}}{90}

. Then

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

\infty

is equal to

Q. If

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

, then find the value of

14 S .

Q. Sum to

n

\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}}

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If 114+124+134+⋯upto ∞=π490, then the value of 114+134+154+⋯upto ∞ is

The correct option is B π496114+134+154+174+… =(114+124+134+144+154+…)−(124+144+164+…) =π490−124(114+124+134+…) =π490−116(π490) =π496

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots upto \infty = \frac{π^{4}}{90},$ then the value of $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \dots upto \infty$ is