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Question Number 117903 by aurpeyz last updated on 14/Oct/20
prove by mathematical induction that n(n+1)(n+2) is an integer multiple of 6
$${prove}\:{by}\:{mathematical}\:{induction} \\ $$$${that}\:{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:{is}\:{an}\:{integer}\: \\ $$$${multiple}\:{of}\:\mathrm{6} \\ $$
Answered by mindispower last updated on 14/Oct/20
(n+1)(n+2)(n+3) =n(n+1)(n+2)+3(n+1)(n+2) =n(n+1)(n+2)+3n^2 +9n+6 =n(n+1)(n+2)+3n(n+1)+6(n+1) p_m =m(m+1)(m+2) 2∣n(n+1) “n=2k tru ,n=2k+1⇒n+1=2(k+1) true” for n=0 we hve 6∣0 true supose ∀n 6∣P_n ;6∣P_(n+1) P_(n+1) =(n+1)(n+2)(n+3)=n(n+1)(n+2)+3n(n+1)+6 p_n =n(n+1)(n+2)=6k by hypothes n(n+1)=2a ⇒p_(n+1) =6(k+a+1)⇒6∣p_(n+1)
$$\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right) \\ $$$$={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)+\mathrm{3}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)+\mathrm{3}{n}^{\mathrm{2}} +\mathrm{9}{n}+\mathrm{6} \\ $$$$={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)+\mathrm{6}\left({n}+\mathrm{1}\right) \\ $$$${p}_{{m}} ={m}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right) \\ $$$$\mathrm{2}\mid{n}\left({n}+\mathrm{1}\right)\:“{n}=\mathrm{2}{k}\:{tru}\:,{n}=\mathrm{2}{k}+\mathrm{1}\Rightarrow{n}+\mathrm{1}=\mathrm{2}\left({k}+\mathrm{1}\right)\:{true}'' \\ $$$${for}\:{n}=\mathrm{0}\:{we}\:{hve}\:\mathrm{6}\mid\mathrm{0}\:{true} \\ $$$${supose}\:\forall{n}\:\:\mathrm{6}\mid{P}_{{n}} \:;\mathrm{6}\mid{P}_{{n}+\mathrm{1}} \\ $$$${P}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)+\mathrm{6} \\ $$$${p}_{{n}} ={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)=\mathrm{6}{k}\:{by}\:{hypothes} \\ $$$${n}\left({n}+\mathrm{1}\right)=\mathrm{2}{a} \\ $$$$\Rightarrow{p}_{{n}+\mathrm{1}} =\mathrm{6}\left({k}+{a}+\mathrm{1}\right)\Rightarrow\mathrm{6}\mid{p}_{{n}+\mathrm{1}} \\ $$$$ \\ $$
Answered by 1549442205PVT last updated on 14/Oct/20
Prove n(n+1)(n+2) is multiply of 6 ∀n∈N,n≥1by mathematical inuction i)For n=1⇒A_1 =1.2.3=6⋮6⇒State is true ii)Suppose State was true ∀n≤k that means A_k =k(k+1)(k+2)⋮6 iii)We will prove that A_(k+1) ⋮6 Indeed,A_(k+1) =(k+1)(k+2)(k+3) =k(k+1)(k+2)+3(k+1)(k+2) =A_k +3(k^2 +3k+2)=A_k +3k(k+1) +6(k+1)⋮6 Since A_k ⋮6,6(k+1)⋮6 3k(k+1)⋮6(by in two sequence numbers always have one even number ) Thus,State is also for n=k+1,so by mathimatical induction principle State is true ∀n∈N^∗ (q.e.d)
$$\mathrm{Prove}\:\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\:\mathrm{is}\:\mathrm{multiply}\:\mathrm{of}\:\mathrm{6} \\ $$$$\forall\mathrm{n}\in\mathrm{N},\mathrm{n}\geqslant\mathrm{1by}\:\mathrm{mathematical}\:\mathrm{inuction} \\ $$$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{n}=\mathrm{1}\Rightarrow\mathrm{A}_{\mathrm{1}} =\mathrm{1}.\mathrm{2}.\mathrm{3}=\mathrm{6}\vdots\mathrm{6}\Rightarrow\mathrm{State}\:\mathrm{is}\:\mathrm{true} \\ $$$$\left.\mathrm{ii}\right)\mathrm{Suppose}\:\mathrm{State}\:\mathrm{was}\:\mathrm{true}\:\forall\mathrm{n}\leqslant\mathrm{k} \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{A}_{\mathrm{k}} =\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)\vdots\mathrm{6} \\ $$$$\left.\mathrm{iii}\right)\mathrm{We}\:\mathrm{will}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{A}_{\mathrm{k}+\mathrm{1}} \vdots\mathrm{6} \\ $$$$\mathrm{Indeed},\mathrm{A}_{\mathrm{k}+\mathrm{1}} =\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)\left(\mathrm{k}+\mathrm{3}\right) \\ $$$$=\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)+\mathrm{3}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right) \\ $$$$=\mathrm{A}_{\mathrm{k}} +\mathrm{3}\left(\mathrm{k}^{\mathrm{2}} +\mathrm{3k}+\mathrm{2}\right)=\mathrm{A}_{\mathrm{k}} +\mathrm{3k}\left(\mathrm{k}+\mathrm{1}\right) \\ $$$$+\mathrm{6}\left(\mathrm{k}+\mathrm{1}\right)\vdots\mathrm{6}\:\mathrm{Since}\:\mathrm{A}_{\mathrm{k}} \vdots\mathrm{6},\mathrm{6}\left(\mathrm{k}+\mathrm{1}\right)\vdots\mathrm{6} \\ $$$$\mathrm{3k}\left(\mathrm{k}+\mathrm{1}\right)\vdots\mathrm{6}\left(\mathrm{by}\:\mathrm{in}\:\mathrm{two}\:\mathrm{sequence}\:\mathrm{numbers}\right. \\ $$$$\left.\mathrm{always}\:\mathrm{have}\:\mathrm{one}\:\mathrm{even}\:\mathrm{number}\:\right) \\ $$$$\mathrm{Thus},\mathrm{State}\:\mathrm{is}\:\mathrm{also}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1},\mathrm{so} \\ $$$$\mathrm{by}\:\mathrm{mathimatical}\:\:\mathrm{induction}\:\mathrm{principle} \\ $$$$\mathrm{State}\:\mathrm{is}\:\mathrm{true}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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