OK, Anonymous, since you forced my hand...
Show that it is true for n = 2....that is
(1 + 2)^2 = 1^3 + 2^3
3^2 = 1 + 8
9 = 9
Assume it is true for k = 1...that is ....
(1 + 2 + 3 +........+ (k-1) + k)^2 = 1^3 + 2^3 + 3^3 + ....+ (k-1)^3 + k^3
Show that it is true for k +1....so....
(1 + 2 + 3 +......+ k + (k+1))^2 = 1+3 + 2^3 + 3^3 +.....+ k^3 + (k+1)^3
And by an identity, the left side can be written as
[(1 + 2 +3 + ......k) + (k+1)]^2 =
(1 + 2 +3 +......k)^2 + 2(k + 1)(1+2 +3 +......+k) + (k +1)^2 =
(1^3 + 2^3 + 3^3 +.....+(k-1)^3 +k^3) + 2( k+ 1)(1+2 +3 +......+k) + (k +1)^2
And setting this = to the right hand side and subtracting like terms from both sides, we have
2(k+1)(1+2+3+....+k) + (k+1)^2 = (k+1)^3 divide through by (k+1)
2(1+2+3+......+k) + (k+1) = (k + 1)^2 subtract (k+1) from each side
2(1 + 2 + 3 +.......+ k) = (k+1)^2 - (k+1) factor the right side
2(1 + 2 + 3 +.......+ k) = (k+1)[(k+1)-1]
2(1 + 2 + 3 +.......+ k) = (k+1)[(k] divide boh sides by 2
(1 + 2 + 3 +........+ k) = (k)(k+1)/2
And the right side is just the "formula" for ths sum of the first k positive integers, i.e, the left side.......
