Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$1^3+2^3+3^3+.......+n^3={\left[\frac{n(n+1)}{2}\right]}^2$

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$1^3+2^3+3^3+.......+n^3={\left(\frac{n(n+1)}{2}\right)}^2$
$P\left(n\right)$:
For $n=1$, we have
$P\left(1\right):1^3=1={\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1\times 2}{2}\right)}^2=1^2=1$, which is true
Let $P\left(k\right)$ be true for some positive integer $k$, i.e.,
$1^3+2^3+3^3+.......+k^3={\left(\frac{k(k+1)}{2}\right)}^2..........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$1^3+2^3+3^3+.......+k^3+{(k+1)}^3$
${\left(\frac{k(k+1)}{2}\right)}^2+{(k+1)}^3$ [Using $\left(i\right)$]
$=\frac{k^2{(k+1)}^2}{4}+{(k+1)}^3$
$=\frac{k^2{(k+1)}^2+4{(k+1)}^3}{4}=\frac{{(k+1)}^2\{k^2+4(k+1\left)\right\}}{4}$
$=\frac{{(k+1)}^2\{k^2+4k+4\}}{4}=\frac{{(k+1)}^2{(k+2)}^2}{4}$
$=\frac{{(k+1)}^2{(k+1+1)}^2}{4}$
$1^3+2^3+3^3+.......+k^3+{k+1}^3={\left(\frac{(k+1)(k+1+1)}{2}\right)}^2$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.