You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:
Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:
1^3 = 1^2(1+1)^2/4 => 1 = 1*4/4 => 1=1
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
P(k): 1^3 + 2^3 + .. + k^3 = (k^2(k+1)^2)/4 holds
P(k+1): 1^3 + 2^3 + .. + k^3 + (k+1)^3 = ((k+1)^2(k+2)^2)/4
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
(k^2(k+1)^2)/4 + (k+1)^3 = ((k+1)^2(k+2)^2)/4
k^2(k+1)^2 + 4(k+1)^3 = (k+1)^2(k+2)^2
Factor out (k+1)^2 to the left side:
(k+1)^2(k^2 + 4k + 4) = (k+1)^2(k+2)^2
Notice that k^2 + 4k + 4 is a perfect square, such that:
k^2 + 4k + 4 = (k+2)^2
(k+1)^2(k+2)^2 = (k+1)^2(k+2)^2
Notice that P(k+1) holds.
Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement P(n):1^3 +2^3 +3^3 + ... + n^3 = (n^2(n+1)^2)/4 holds for all positive integers n.
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