Question

By the principle of mathematical induction, prove the following:

1³ + 2³ + 3³ + ….. + n³ = ${\displaystyle \frac{{\text{n}}^2{\left(\text{n + 1}\right)}^2}{4}}$ for all x ∈ N.

Answer 1

Let P(n) be the statement 1³ + 2³ + 3³ + ….. + n³ = ${\displaystyle \frac{{\text{n}}^2{\left(\text{n + 1}\right)}^2}{4}}$ for all x ∈ N.

i.e., p(n) = 1³ + 2³ + …… + n³ = ${\displaystyle \frac{{\text{n}}^2{\left(\text{n + 1}\right)}^2}{4}}$ for all x ∈ N

Put n = 1

LHS = 1³ = 1

RHS = ${\displaystyle \frac{1^2{(1+1)}^2}{4}}$

${\displaystyle =\frac{1\times 2^2}{4}}$

${\displaystyle =\frac{4}{4}}$ = 1

∴ P(1) is true.

Assume that P(n) is true n = k

P(k): 1³ + 2³ + …… + k³ = ${\displaystyle \frac{k^2{(k+1)}^2}{4}}$

To prove P(k + 1) is true.

i.e., to prove 1³ + 2³ + ……. + k³ + (k + 1)³ = ${\displaystyle \frac{{(k+1)}^2{((k+1)+1)}^2}{4}=\frac{{(k+1)}^2{(k+2)}^2}{4}}$

Consider 1³ + 2³ + …… + k³ + (k + 1)³ = ${\displaystyle \frac{k^2{(k+1)}^2}{4}+{(k+1)}^3}$

= (k + 1)² ${\displaystyle [\frac{k^2}{4}+(k+1)]}$

= (k + 1)² ${\displaystyle \left[\frac{k^2+4(k+1)}{4}\right]}$

${\displaystyle =\frac{{(k+1)}^2{(k+2)}^2}{4}}$

⇒ P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction P(n) is true for all n ∈ N.