Proof of ∑(x^2)

First we will prove ∑x

x^2-(x-1)^2=2x-1

taking summation on both sides .x ranging from 0 to n

∑(x^2)-∑((x-1)^2)=2∑x -∑(1)

∑(x^2)-(∑((x)^2)-n^2)=2∑x - n

∑(x^2)-∑(x^2)+n^2=2∑x- n

n^2=2∑x +n

2∑x =n^2+n

∑x =n(n+1)/2

proof for ∑x^2

consider x^3-(x-1)^3=3.x^2-3.x+1

Take summation on both sides from 1 to n

∑(x^3)-∑((x-1)^3)=3∑x^2 -3∑x + ∑1

∑(x^3)-(∑(x^3)-n^3)=3∑x^2 –3n(n+1)/2 + n

n^3=3∑x^2 –3n(n+1)/2 + n

3∑x^2 =n.n.n+3n(n+1)/2 - n

3∑x^2 =n(2* (n^2)+3*(n+1)-2)/2

3∑x^2 =n(2*(n^2)+3*n+1)/2

3∑x^2 =n(n+1)(2n+1)/2

∑x^2 = n(n+1)(2n+1)/6

ps: take each sigma ranging from 0 to n