Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution:

Let the given statement be $\mathrm{P}(n)$, i.e.,

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

For $n=1$, we have

$P(1): 1^{3}=1=\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{1.2}{2}\right)^{2}=1^{2}=1$, which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

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

We shall now prove that $\mathrm{P}(k+1)$ is true.

Consider

$1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}$

$=\left(\frac{k(k+1)}{2}\right)^{2}+(k+1)^{3}$ [Using (i)]

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

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

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

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

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

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

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

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

Thus, $\mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true.

Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.