Question

Use the principle of mathermatical induction to prove that
$1^3+2^3+.....n^3=\frac{1}{4}{n}^2{(n+1)}^2$ for every natural number $n$.

Answer 1

Let $P\left(n\right):1^3+2^3+3^3+......+n^3={\left(\frac{n(n+1)}{2}\right)}^2$

For $n=1$

$LHS=1^3=1$

$RHS={\left(\frac{1(1+1)}{2}\right)}^2=1$

Hence $LHS=RHS$

$\therefore P\left(n\right)$ is true for $n=1$

Assume that P(k) is true

$1^3+2^3+3^3+....+k^3={\left(\frac{k(k+1)}{2}\right)}^2......\left(1\right)$

We will prove that P(k+1) is true

$1^3+2^3+3^3+....+k^3={\left(\frac{(k+1)(k+1+1)}{2}\right)}^2$

$1^3+2^3+3^3+....+k^3={\left(\frac{(k+1)(k+2)}{2}\right)}^2......\left(2\right)$

We have to prove P(k+1) from P(k) i.e (2) from (1)

From (1)

$1^3+2^3+3^3+.....+k^3={\left(\frac{k(k+1)}{2}\right)}^2$

Adding $(k+1{)}^3$ both sides,

$1^3+2^3+3^3+.....+k^3+(k+1{)}^3={\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3$

$=\frac{k^2(k+1{)}^2}{4}+(k+1{)}^3$

$=\frac{k^2(k+1{)}^2+4(k+1{)}^3}{4}$

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

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

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

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

$={\left(\frac{(k+1)(k+2)}{2}\right)}^2$

Thus, $1^3+2^3+3^3+....+k^3={\left(\frac{(k+1)(k+2)}{2}\right)}^2$ i.e P(k+1) is true whenever P(k) is true.

$\therefore$ By the principle of mathematical induction, P(n) is true for n, where n is a natural number.