Question

A question of sequence and series pls ans

Answer 1

where is it?

Answer 2

I am writing down the equation

Answer 3

if \[1/1^{2} + 1/2^{2} +1/3^{2}.......upto{\infty}, then \frac{ 1}{ 2^{2}}+\frac{ 1}{ 4^{2}}\]

Answer 4

\[\sum_{n=1}^{\infty}\frac{ 1 }{ n^2 }\]

Answer 5

u want to calculate this?

Answer 6

upto infinity value is \[\pi^{2}/6\]

Answer 7

I want to calculate valueof\[\sum_{n=1}^{\infty}1/(2n)^2\]

Answer 8

@ganeshie8 help

Answer 9

\[\frac{ \pi^2 }{ 24 }?\]

Answer 10

no

Answer 11

\[\pi^2/8\]

Answer 12

\[\frac{ 1 }{ 1^2 }+\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 3^2 }+...=\frac{ \pi^2 }{ 6 }\]

Answer 13

is this is the info given?

Answer 14

yes

Answer 15

\[\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 4^2 }+\frac{ 1 }{ 6^2 }+....\]

Answer 16

we have to find this?

Answer 17

yes

Answer 18

then the answer should be pi^2/24

Answer 19

\[\frac{ 1 }{ 4 }(1+\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 3^2 }+......)\]

Answer 20

\[\frac{ 1 }{ 4 }\frac{ \pi^2 }{ 6 }\]

Answer 21

i think u want \[\frac{ 1 }{ 1^2 }+\frac{ 1 }{ 3^2 }+\frac{ 1 }{ 5^2 }+....\]

Answer 22

yes can u find \[\frac{ 1}{ 1^2}+\frac{1}{3^2} +\frac{1}{5^2}\].......

Answer 23

then that would be\[\frac{ \pi^2 }{ 6 }-\frac{ \pi^2 }{ 24 }=\frac{ \pi^2 }{ 8 }\]

Answer 24

ohhh thx

Answer 25

@kanwal32 please be clear with your question

Answer 26

ok

Answer 27

:)