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1/9 + 1/16 + 1/25 + 1/36 + ......... I know it has something to do with a square in the denominator..
1/n^2 , from 3-infinity
\[\sum_{1}^{\infty}1/(n+2)^{2}\]

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Other answers:

\[\sum_{n=3}^{\infty} \frac{1}{n^2} \]
yeah
which one?
both are same
oh okay thank you :)
what about for 1 - 1/3 + 1/5 - 1/7 +.......
\[\sum_{n=0}^{\infty} (-1)^n \frac{1}{2n+1}\]
and 125/9 + 625/16 + 3,125/25 + 15,625/36 . I know the bottom is n^2 but what's the top?
the top are factgorials im fairly sure
5!, 6! etc
\[\sum_{n=3}^{ \infty} \frac{ (n+2)!}{n^2} \]
wait, thats not right :|
yeah
its multiples of 5 on the top
its powers of 5
5^n / n^2
thats it
n=3 to infinity
\[\sum_{n=3}^{\infty} 5^n \frac{5^2}{n^2} \]
\[\sum_{n=3}^{\infty} \frac{5^{n+2}}{n^2}\]
ur wrong mate
if it starts from 3 then its 5^n
no, thats not it :|
yeh whatever
?
\[\sum_{n=3}^{\infty}5^{n}/n ^{2}\]
thats ur answer
you're right ! Thanks :D Saved me
no prob...
hey could you look at myt problem pleaseeee

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