Question-190129

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

Answered by ARUNG_Brandon_MBU last updated on 28/Mar/23

Let P_n be the statement 3∣ n^3 +2n ∀n≥1 P_n is true for n=1 , n=2 Suppose P_n true for n and prove that it′s also true for n+1. 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∣(n^3 +2n) from hypothesis and 3∣3(n^2 +n+1) thus P_(n+1) is true.

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

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Question-190129

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