If 1-1-4-1-9-1-16- 2-6 then the value of 1-4-1-16-1-36-1-64- is equals

The correct option is D π ^2/24 #160; #160; #160; #160; Sum = #160; 14 + 1/16 + 1/36 + 1/64 + ......... ...

149

You visited us 149 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If 1+1/4+1/...

Question

If $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . = \frac{π^{2}}{6}$ then the value of $\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + . . .$ is equals

\frac{π^{2}}{6}

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

\frac{π^{2}}{12}

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

\frac{π^{2}}{8}

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

\frac{π^{2}}{24}

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

Open in App

Solution

The correct option is D $\frac{π^{2}}{24}$
$S u m$ $=$ $\frac{1}{4}$ + $\frac{1}{16}$ + $\frac{1}{36}$ + $\frac{1}{64}$ + .........
Taking $\frac{1}{4}$ common, we get
$=$ $\frac{1}{4}$ ( 1 + $\frac{1}{4}$ + $\frac{1}{9}$ + $\frac{1}{16}$ + .........)
$=$ $\frac{1}{4}$ . $\frac{π^{2}}{6}$ $=$ $\frac{π^{2}}{24}$

Suggest Corrections

Similar questions

I f \sum_{r = 1}^{\infty} \frac{1}{(2 r - 1)^{2}} = \frac{π^{2}}{8}

then the value of x =

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

Q. The maximum value of the function

f (x) = \int_{0}^{1} t sin (x + π t) d t, x \in R

is -

Q. The maximum value of the function

f (x) = 1 \int 0 t sin (x + π t) d t, x \in R

Q. The value of

\frac{π}{2} \int - \frac{π}{2} \frac{x^{2} cos x}{1 + e^{x}} d x

is equal to

If $\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . \infty = \frac{π^{2}}{6}$ , then $\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} +$ .... equals

Join BYJU'S Learning Program

If 1+14+19+116+...=π26 then the value of 14+116+136+164+... is equals

The correct option is D π224 Sum = 14 + 116 + 136 + 164 + ......... Taking 14 common, we get = 14 ( 1 + 14 + 19 + 116 + .........) = 14. π26 = π224

If $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . = \frac{π^{2}}{6}$ then the value of $\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + . . .$ is equals