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The Brainliest Answer!
2015-05-11T16:49:36+05:30
Recall the fact that the sum of first k odd natural numbers equals k^2
1+3+5+\cdots+(2k-1)=k^2

That means the general term of the series is simply k^2

1+(1+3)+(1+3+5)+\cdots+(1+3+5+\cdots+(2n-1))

=1+(2^2)+(3^2)+\cdots+(n^2)

=\sum\limits_{k=1}^n~k^2

=\boxed{\dfrac{n(n+1)(2n+1)}{6}}
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2015-05-11T19:49:11+05:30
1 + (1+3)+(1+3+5)+..... = this will be something like : 1+4+9+16+25+36.... ....  (ie. the sum of squares of natural numbers)

and the sum of squares of n terms can be given by : n(n+1)(2n+1)/6
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