11+1/32+1/52......
=Σ1/(2n+1)2
=Ï€2/8
i think i found this in d targetiit formula sheet.........
If 1+\frac{1}{4}+\frac{1}{9}+....inf is equal to \frac{\pi ^{2}}{6}
then 1+\frac{1}{9}+\frac{1}{25} .....inf
is equal to
a) \frac{\pi ^{2}}{8}
b)\frac{\pi ^{2}}{3}
c)\frac{\pi ^{2}}{12}
d)\frac{\pi ^{2}}{4}
how do you solve this . plz help kal xam hai .
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9 Answers
1
1i dunno...........
by d way....i dint give any method over there hehe...so ur "thanks" was a waste hehe
1hey u can do that .i mean u can solve that.
1+1/4+1/9+1/16+1/25+...= 1+1/4+1/16+1/36+.......+ 1/9 +1/ 25+1/49+....=pi2 /6
and the first one is a standard sequence. so solve it
and then subtract it form pi2 /6
and u ll get the answer.
Try it.
1u dont need to kno no formula for dis one...
pi2/24= 1/4 +1/16+1/36.... (from the first one, dividing by 4 both sides)
and anyways pi2/6=1+1/4+1/9......
subtracting the 2nd from 1st,
we have the req.d series as pi2/8...
cheers!!!
62There is a better way
S=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}....
Thus, S=1+\frac{1}{9}+\frac{1}{25}....\\+\frac{1}{4}{1+\frac{1}{4}+\frac{1}{9}+.. inf}
Thus, S=1+\frac{1}{9}+\frac{1}{25}....+\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+
Thus, S=1+\frac{1}{9}+\frac{1}{25}....+\frac{1}{4}\left\{{1+\frac{1}{4}+\frac{1}{9}+.. inf} \right\}
S=1+\frac{1}{9}+\frac{1}{25}....+\left\{\frac{1}{4}S \right\}
1+\frac{1}{9}+\frac{1}{25}....+...inf=\frac{\pi^2}{8}
62hmm.. good work gordo..
I din realise that you have already solved this one :)