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Question Number 137875 by physicstutes last updated on 07/Apr/21

$$\mathrm{Proof}\mathrm{by}\mathrm{mathematical}\mathrm{induction}\mathrm{that}$$$$f\left(n\right)=n^3+5n$$$$\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}6.$$

Answered by mathmax by abdo last updated on 08/Apr/21

$$\mathrm{f}\left(0\right)=0\mathrm{is}\mathrm{mulyiple}\mathrm{of}6\mathrm{let}\mathrm{suppose}\mathrm{f}\left(\mathrm{n}\right)\mathrm{multiple}\mathrm{of}6\Rightarrow$$$$\mathrm{f}\left(\mathrm{n}\right)=6\mathrm{k}\Rightarrow {\mathrm{n}}^3+5\mathrm{n}=6\mathrm{k}$$$$\mathrm{f}(\mathrm{n}+1)={(\mathrm{n}+1)}^3+5(\mathrm{n}+1)={\mathrm{n}}^3+{3\mathrm{n}}^2+3\mathrm{n}+1+5\mathrm{n}+5$$$$={\mathrm{n}}^3+5\mathrm{n}+{3\mathrm{n}}^2+3\mathrm{n}+6=6\mathrm{k}+3\mathrm{n}(\mathrm{n}+1)+6\mathrm{but}\mathrm{n}(\mathrm{n}+1)\mathrm{is}\mathrm{multiple}$$$$\mathrm{of}2\Rightarrow \mathrm{n}(\mathrm{n}+1)=2\mathrm{q}\Rightarrow \mathrm{f}(\mathrm{n}+1)=6\mathrm{k}+6+6\mathrm{q}=6(\mathrm{k}+\mathrm{q}+1)\Rightarrow$$$$\mathrm{f}(\mathrm{n}+1)\mathrm{is}\mathrm{multiple}\mathrm{of}6$$

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