(1)+(1+2)+(1+2+3)....... it is in a A.P series.Find sumtotal upto n-th term.

This is strictly NOT a A.P. This is called "Sequence of trianangular numbers", and can be summed up using the following formula: n[n +1]*[n + 2] / 6. So, the first 10 terms will sum up to:[10*11*12]/6=220.

I think that the sumtotal upto n is n² .

edit: never mind it is wrong, I only tried when n=1,2,3 and all of them had a sumtotal of a squared number.