1+(2+3)+(4+5+6)+..... to n terms

Hi

this is simply the formula for sum of first n natural numbers  and its value is n(n+1)/2

Regards

Hi Keshaw,

The question simply reduces to finding the number of terms in the series.

1+2+3+.....n = n(n+1)/2 terms are there in the series.

Hence the sum of the first n(n+1)/2 terms will be n(n+1)[n(n+1)+2]/4 = n(n+1)(n²+n+2)/4------- (this is the answer).

Hope it helps.

Regards,

Ashwin (IIT MadraS).

The nth bracket contains n terms.The first term of the nth bracket is a number of the series 1+2+4+…...+nLet a be that term.a=((n^2)-n+2)÷2Sum of each bracket=(n/2)[2{n^2-n+2}/2+(n-1)×1]=(n/2){(n^2)+1}Putting n=1,2,3...nS=(1/2)[(1^2+2^2+3^2....+n^2)+(1+2+3+4...n)]S=n(n+1)(n^2+n+2)/8Hope this answer helps.