Question

The sum of n terms of the series 1 + (1+3)+(1+3+5)+..... is?

Answer 1

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 $k^2$

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

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

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

$=\boxed{\dfrac{n(n+1)(2n+1)}{6}}$

Answer 2

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