(1)+(2+3)+(4+5+6)......n thterm, find out the sum of this series upto n th term.

Note that the last number in the bracket of each term (1,3,6,...) is simply the set of triangular numbers. So, in nth bracket, the last term would be n(n+1)/2 [nth triangular number].

Now, the sum of the given series is clearly equal to S= 1+2+3+.....+m where m is the last term of the nth bracket.

So S= m(m+1)/2 (sum of natural numbers)

But m= n(n+1)/2

So substitute this value of m in S to get

S= n(n+1)(n^2+n+2)/8