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