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Question Number 190129 by yaslm last updated on 27/Mar/23

Answered by ARUNG_Brandon_MBU last updated on 28/Mar/23

$$\mathrm{Let}{\mathrm{P}}_n\mathrm{be}\mathrm{the}\mathrm{statement}3\mid n^3+2n\mathrm{\forall }n⩾1$$$${\mathrm{P}}_n\mathrm{is}\mathrm{true}\mathrm{for}n=1,n=2$$$$\mathrm{Suppose}{\mathrm{P}}_n\mathrm{true}\mathrm{for}n\mathrm{and}\mathrm{prove}\mathrm{that}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{also}\mathrm{true}\mathrm{for}n+1.$$$${\mathrm{P}}_{n+1}={(n+1)}^3+2(n+1)$$$$=n^3+3n^2+3n+1+2n+2$$$$=(n^3+2n)+(3n^2+3n+3)$$$$=(n^3+2n)+3(n^2+n+1)$$$$3\mid (n^3+2n)\mathrm{from}\mathrm{hypothesis}\mathrm{and}3\mid 3(n^2+n+1)$$$$\mathrm{thus}{\mathrm{P}}_{n+1}\mathrm{is}\mathrm{true}.$$

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