1+(1+2)+(1+2+3)+(1+2+3+4)... find the sum of the series
In order to find the sum of this series, remember two basic formulas
Sum of first n natural numbers
1 + 2 + 3 + .....+ n
∑_{ (1 to n)} {n} = (1/2)n(n + 1)

Sum of first n squares
1^{2} + 2^{2} + 3^{2} + ... + n^{2}
∑ _{(1 to n)} {n^{2}} = (1/6) n(n + 1)(2n + 1)

Now let's arrange the series
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + .......................(1 + 2 + 3 + 4 + .........+ n)
∑ _{(1 to n)} { 1 + 2 + 3 + 4 + ......+..n }
∑ _{(1 to n)} {(1/2) n(n+1)}
∑ _{(1 to n)} (1/2){n^{2} + n}
(1/2) ∑ _{(1 to n)} {n^{2} + n}
(1/2) ∑_{ (1 to n)} {n^{2}} + (1/2) ∑ _{(1 to n)} {n}
(1/2)(1/6) n(n + 1)(2n + 1) + (1/2)(1/2)n(n + 1)
(1/2)n(n + 1) *[{(1/6)(2n + 1)} + (1/2)]
(1/2)n(n + 1) *[(1/6)(2n + 4)]
= (1/6) n(n + 1)(n + 2)
This is the sum upto n terms.