If 114-124-134-upto -490- then the value of 114-134-154-upto - is

11^4 + 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 12^4 ( 11^4+12^4 + 13^4 + nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Infinite GP

If 114+124+13...

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{4}}{96}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π^{4}}{124}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{4}}{196}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. If

y = 1 + \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{3}} + \dots

\infty

with

| x | > 1

, then

\frac{d y}{d x} =

Q. The value of

\frac{1}{{log}_{3} 60} + \frac{1}{{log}_{4} 60} + \frac{1}{{log}_{5} 60},

y = x^{2} + \frac{1}{x^{2} + \frac{1}{\frac{1}{x^{2} + \dots}}}

, then

\frac{d y}{d x}

Q. The value of

\frac{1}{cos 290^{\circ}} + \frac{1}{\sqrt{3} sin 250^{\circ}}

\frac{1}{2 x} + \frac{1}{x} = \frac{3}{2}

Join BYJU'S Learning Program

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