find the sum of the series:-1+(1+3)+(1+3+5).............n terms

In this series the nth term is given by 

(1+3+5+.................r) Terms which is an AP of first term 1 and cd 2 therefor sum upto r is r/2[2+(r-1)2] = r^2

Hence of series is Sigma(r^2) = n(n+1)(2n+1)/6

after simplification, we see that it is just the same as n^2.

Sum=n(n+1)(2n+6)/6