Question Number 6350 by sanusihammed last updated on 24/Jun/16
![Prove by matimatical induction 1 + 8 + 27 + ...... + n^3 = [((n(n + 1))/2)]^2](https://www.tinkutara.com/question/Q6350.png)
$${Prove}\:{by}\:{matimatical}\:{induction} \\ $$$$\mathrm{1}\:+\:\mathrm{8}\:+\:\mathrm{27}\:+\:……\:+\:{n}^{\mathrm{3}} \:\:=\:\:\left[\frac{{n}\left({n}\:+\:\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$
Answered by prakash jain last updated on 24/Jun/16
![n=1 1=(((1∙2)/2))^2 so statement is true for n=1 assume it is true for n=k then Σ_(i=1) ^k i^3 =[((k(k+1))/2)]^2 Σ_(i=1) ^(k+1) i^3 =[((k(k+1))/2)]^2 +(k+1)^3 =(k+1)^2 [(k^2 /2^2 )+k+1] =(((k+1)^2 )/2^2 )[k^2 +4k+k)=(((k+1)^2 (k+2)^2 )/2^2 ) if statement is true for n=k then it is true for n=k+1. ■](https://www.tinkutara.com/question/Q6356.png)
$${n}=\mathrm{1} \\ $$$$\mathrm{1}=\left(\frac{\mathrm{1}\centerdot\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} \:\mathrm{so}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{1} \\ $$$${assume}\:{it}\:{is}\:{true}\:{for}\:{n}={k} \\ $$$$\mathrm{then}\:\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}^{\mathrm{3}} =\left[\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}{i}^{\mathrm{3}} =\left[\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{3}} =\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left[\frac{{k}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+{k}+\mathrm{1}\right] \\ $$$$=\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }\left[{k}^{\mathrm{2}} +\mathrm{4}{k}+{k}\right)=\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} } \\ $$$${if}\:{statement}\:{is}\:{true}\:{for}\:{n}={k}\:{then}\:{it}\:{is}\:{true}\:{for} \\ $$$${n}={k}+\mathrm{1}. \\ $$$$\blacksquare \\ $$
Commented by sanusihammed last updated on 25/Jun/16
$${Thanks}\:{so}\:{much} \\ $$