Introduction to P.M.I
Question

# The identity ${1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{n}}^{3}$ is equal to

Easy
Solution

## Let the given statement be P(n).$\mathrm{P}\left(\mathrm{n}\right):{1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{n}}^{3}={\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]}^{2}$Step I : For n=1,             $\mathrm{P}\left(1\right):{\left[\frac{1\left(1+1\right)}{2}\right]}^{2}={\left[\frac{1×2}{2}\right]}^{2}={1}^{2}=1={1}^{3}$ which is trueStep ll: Let it is true for n = k,             ${1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{k}}^{3}={\left[\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right]}^{2}------\left(\mathrm{i}\right)$Step lll: For n=k+1,             $\left({1}^{3}+{2}^{3}+{3}^{3}+{4}^{3}+\dots +{\mathrm{k}}^{3}\right)+\left(\mathrm{k}+1{\right)}^{3}$           $={\left[\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right]}^{2}+\left(\mathrm{k}+1{\right)}^{3}$      [using Eq. (i)]            $\begin{array}{r}=\frac{{\mathrm{k}}^{2}\left(\mathrm{k}+1{\right)}^{2}}{4}+\frac{\left(\mathrm{k}+1{\right)}^{3}}{1}\\ =\frac{{\mathrm{k}}^{2}\left(\mathrm{k}+1{\right)}^{2}+4\left(\mathrm{k}+1{\right)}^{3}}{4}\end{array}$  On taking (k + 1)2 common in numerator Part,            $\begin{array}{l}=\frac{\left(\mathrm{k}+1{\right)}^{2}\left[{\mathrm{k}}^{2}+4\left(\mathrm{k}+1\right)\right]}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left({\mathrm{k}}^{2}+4\mathrm{k}+4\right)}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left(\mathrm{k}+2{\right)}^{2}}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left[\left(\mathrm{k}+1\right)+1{\right]}^{2}}{4}\\ ={\left[\frac{\left(\mathrm{k}+1\right)\left\{\left(\mathrm{k}+1\right)+1\right\}}{2}\right]}^{2}\end{array}$Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

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