Question

Prove the following using mathematical induction.

1. The sum of the cubes of the first n integers:
a) Basis Step
b) Inductive Step

Answer 1

I assume the claim is being made for integers
`n\ge1`, not just any integer.

Basis step: Let
`n=1`. Then

`\displaystyle \sum_(i=1)^1 i^3 = 1^3 = 1`

`\frac{1^2(1+1)^2}4 = \frac{2^2}4 = 1`

so this case holds.

Inductive step: Assume

`\displaystyle \sum_(i=1)^k i^3 = \frac{k^2 (k+1)^2}4`

As a result, we want to show that

`\displaystyle \sum_(i=1)^(k+1) i^3 = \frac{(k+1)^2 (k+2)^2}4`

Now,

`\displaystyle \sum_(i=1)^(k+1) i^3 = \sum_(i=1)^k i^3 + (k+1)^3 \\\\ ~~~~~~~~ = \frac{k^2(k+1)^2}4 + (k+1)^3 \\\\ ~~~~~~~~ = \frac{k^2(k+1)^2 + 4(k+1)^3}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 \bigg(k^2 + 4(k+1)\bigg)}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 (k^2 + 4k + 4)}4 \\\\ ~~~~~~~~ = \frac{(k+1)^2 (k+2)^2}4`

QED