Hi Angels,
The subject line of your email was Mathematical Induction so I assume that is how you want to solve this problem. There are two steps to proving this fact using mathematical induction. The first step is to show that the result is true when n = 1. The second step is to show that whenever the result is true for a value of n, say n = k, it is also true for n = k + 1.
Step 1.
When n = 1 the left side is 1^{3} = 1 and the right side is [1^{2} (1 + 1)^{2]}/4 = 2^{2}/4 = 1. Thus the result is true when n = 1.
Step 2.
Suppose that the result is true for n = k, that is assume that
1^{3} + 2^{3} + 3^{3} + 4^{3} + ··· + k^{3} = [k^{2} (k + 1)^{2}]/4
You are to prove that the result is true for n = k + 1, that is you are to prove that
1^{3} + 2^{3} + 3^{3} + 4^{3} + ··· + k^{3} + (k + 1)^{3} = [(k + 1)^{2} (k + 1 + 1)^{2}]/4 = [(k + 1)^{2} (k + 2)^{2}]/4
Since you are assuming that 1^{3} + 2^{3} + 3^{3} + 4^{3} + ··· + k^{3} = [k^{2} (k + 1)^{2}]/4,
1^{3} + 2^{3} + 3^{3} + 4^{3} + ··· + k^{3} + (k + 1)^{3} = [k^{2} (k + 1)^{2}]/4 + (k + 1)^{3}
Simplify this expression and show it is equal to [(k + 1)^{2} (k + 2)^{2}]/4
I hope this helps,
Harley
