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)

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Sum of first n squares

1² + 2² + 3² + ... + n²

∑ _{(1 to n)} {n²} = (1/6) n(n + 1)(2n + 1)

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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){n² + n}

(1/2) ∑ _{(1 to n)} {n² + n}

(1/2) ∑ _{(1 to n)} {n²} + (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.