Question

Prove that for every natural number n.

1³ + 2³ + 3³ + ... + n³ = \(\big[\frac{n(n+1)}{2}\big]^2\)

Answer 1

Let T(n) be the statement, 

1³ + 2³ + 3³ + ... + n³ = \(\frac{n^2(n+1)^2}{4}\)

Basic Step: For n = 1, 1³ = \(\frac{1^2(1+1)^2}{4}\) = 1 ⇒ T(1) is true. 

Induction Step: Let T(k) hold for a natural number k, that is 

1³ + 2³ + 3³ + ... + k³ = \(\frac{k^2(k+1)^2}{4}\)

Now, to obtain T(k + 1), add the (k + 1)^th term = (k + 1)³ to both the sides of T(k), i.e., 

1³ + 2³ + 3³ + ... + k³ + (k + 1)³ = \(\frac{k^2(k+1)^2}{4}\) + (k+1)³  

 \(=\frac{(k+1)^2}{4}\) [k²+4(k+1)] = \(\frac{(k+1)^2\,(k+2)^2}{4}\) = \(\big[\frac{(k+1)(k+2)}{2}\big]^2\)

Hence T (k + 1) is true, whenever T(k) is true.