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

Proof by mathematical induction that f(n) = n^3 + 5n is a multiple of 6.

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

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

f(0)=0 is mulyiple of 6 let suppose f(n) multiple of 6 ⇒ f(n)=6k ⇒n^3 +5n =6k f(n+1)=(n+1)^3 +5(n+1) =n^3 +3n^2 +3n+1 +5n +5 =n^3 +5n +3n^2 +3n +6 =6k +3n(n+1) +6 but n(n+1)is multiple of 2 ⇒n(n+1)=2q ⇒f(n+1)=6k+6+6q =6(k+q+1) ⇒ f(n+1) is multiple of 6

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

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