Maths Induction

Let S(n) be the statement
1^3 + 2^3 + 3^3 + .... + n^3 = (1 + 2 + 3 + ..... + n)^2
for all positive integers n

when n=1,
1^3=1=1^2
so, S(1) is true

Assume that S(k) is true for some positive integers k
i.e. 1^3 + 2^3 + 3^3 + .... + k^3 = (1 + 2 + 3 + ..... + k)^2

when n=k+1,
1^3 + 2^3 + 3^3 + .... + k^3 + (k+1)^3
=(1 + 2 + 3 + ..... + k)^2 + (k+1)^3
=(1 + 2 + 3 + ..... + k)^2 + (k+1) (k+1) (k+1)
=(1 + 2 + 3 + ..... + k)^2 + (k+1) [(k+1)k + (k+1)]
=(1 + 2 + 3 + ..... + k)^2 + (k+1) {[(k+1) + (k+1) + (k+1) + ..... + (k+1)] + (k+1)}
(where (k+1)k = summation (k+1) by k times)
=(1 + 2 + 3 + ..... + k)^2
+ (k+1) { [(k+1) + (k-1+2) + (k-2+3) + ..... + (1+k)] + (k+1)}
=(1 + 2 + 3 + ..... + k)^2
+ (k+1) { [k + (k-1) + (k-2) + ..... + 1] + (1 + 2 + 3 + ..... + k) + (k+1) }
=(1 + 2 + 3 + ..... + k)^2
+ (k+1) [ (1 + 2 + 3 + ..... + k) + (1 + 2 + 3 + ..... + k) + (k+1) ]
=(1 + 2 + 3 + ..... + k)^2 + (k+1) [ 2 (1 + 2 + 3 + ..... + k) + (k+1) ]
=(1 + 2 + 3 + ..... + k)^2 + 2 (k+1) (1 + 2 + 3 + ..... + k) } + (k+1)^2
=[(1 + 2 + 3 + ..... + k) + (k+1)]^2
=[1 + 2 + 3 + ..... + (k+1)]^2
so, S(K+1) is true

by principle of M.I.,
1^3 + 2^3 + 3^3 + .... + n^3 = (1 + 2 + 3 + ..... + n)^2
for all positive integers n

2013-06-20 09:00:38 補充：
我的答案沒有用1 + 2 + 3 + ... + n = n(n + 1)/2

回答者編號001,
Using the relation 1 + 2 + 3 + ... + n = n(n + 1)/2 is NOT acceptable.

= (1³ + 2³ + 3³ + ··· + n³) + (n + 1)(n + n² + n + 1) -- using relation "1 + 2 + 3 + ··· + n = (1 + n)(n /2)", i.e.(2 + 4 + 6 + ··· + 2n) = 2(1 + 2 + 3 + ··· + n)
這個不可以用.