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Question Number 66102 by Rio Michael last updated on 09/Aug/19

prove by mathematical induction that 4^n +3^n +2 is a multiple of 3 for all positive integral values of n.

$${prove}\:{by}\:{mathematical}\:{induction}\:{that}\:\:\mathrm{4}^{{n}} +\mathrm{3}^{{n}} +\mathrm{2}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{3}\:{for}\:{all}\: \\ $$$${positive}\:{integral}\:{values}\:{of}\:{n}. \\ $$

Commented by mathmax by abdo last updated on 09/Aug/19

let A_n =4^n +3^n +2 A_1 =9 and 9 multiple of 3 reltion true let suppose A_n multiple of 3 ⇒ A_n =3k k integr on A_(n+1) =4^(n+1) +3^(n+1) +2 =4.(A_n −3^n −2)+3^(n+1) +2 =4A_n −4.3^n −8 +3^(n+1) +2 =4 A_n −3^n −6 =4(3k)−3^n −6 =3{4k−3^(n−1) −2}=3k^′ ⇒ A_(n+1) is multiple of 3 the relation is true for (n+1).

$${let}\:{A}_{{n}} =\mathrm{4}^{{n}} \:+\mathrm{3}^{{n}} \:+\mathrm{2} \\ $$$${A}_{\mathrm{1}} =\mathrm{9}\:{and}\:\:\mathrm{9}\:{multiple}\:{of}\:\mathrm{3}\:\:{reltion}\:{true}\:{let}\:{suppose}\:{A}_{{n}} \:{multiple} \\ $$$${of}\:\mathrm{3}\:\Rightarrow\:{A}_{{n}} =\mathrm{3}{k}\:\:\:{k}\:{integr}\:{on}\:{A}_{{n}+\mathrm{1}} =\mathrm{4}^{{n}+\mathrm{1}} \:+\mathrm{3}^{{n}+\mathrm{1}} \:+\mathrm{2} \\ $$$$=\mathrm{4}.\left({A}_{{n}} −\mathrm{3}^{{n}} −\mathrm{2}\right)+\mathrm{3}^{{n}+\mathrm{1}} \:+\mathrm{2}\:=\mathrm{4}{A}_{{n}} −\mathrm{4}.\mathrm{3}^{{n}} \:−\mathrm{8}\:+\mathrm{3}^{{n}+\mathrm{1}} \:+\mathrm{2} \\ $$$$=\mathrm{4}\:{A}_{{n}} −\mathrm{3}^{{n}} \:−\mathrm{6}\:=\mathrm{4}\left(\mathrm{3}{k}\right)−\mathrm{3}^{{n}} −\mathrm{6}\:=\mathrm{3}\left\{\mathrm{4}{k}−\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{2}\right\}=\mathrm{3}{k}^{'} \:\Rightarrow \\ $$$${A}_{{n}+\mathrm{1}} {is}\:{multiple}\:{of}\:\mathrm{3}\:\:{the}\:{relation}\:{is}\:{true}\:{for}\:\left({n}+\mathrm{1}\right). \\ $$

Commented by Prithwish sen last updated on 11/Aug/19

you have to prove it for n=1 and 2 by putting n=1 and n=2 then assume it is true for n=k i.e 4^k + 3^k +2 is a multiple of 3 now for n=k+1 4^k .4+3^k .3+1=4(4^k +3^k +1) −(3^k +3) ∵ this is multiple of 3 ∴ it is proved for n=k+1

$$\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{for}\:\mathrm{n}=\mathrm{1}\:\mathrm{and}\:\mathrm{2}\:\mathrm{by}\:\mathrm{putting}\: \\ $$$$\mathrm{n}=\mathrm{1}\:\mathrm{and}\:\mathrm{n}=\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{assume}\:\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}\: \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{4}^{\mathrm{k}} +\:\mathrm{3}^{\mathrm{k}} +\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}\: \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$\mathrm{4}^{\mathrm{k}} .\mathrm{4}+\mathrm{3}^{\mathrm{k}} .\mathrm{3}+\mathrm{1}=\mathrm{4}\left(\mathrm{4}^{\mathrm{k}} +\mathrm{3}^{\mathrm{k}} +\mathrm{1}\right)\:−\left(\mathrm{3}^{\mathrm{k}} +\mathrm{3}\right) \\ $$$$\because\:\mathrm{this}\:\:\mathrm{is}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}\: \\ $$$$\therefore\:\mathrm{it}\:\mathrm{is}\:\mathrm{proved}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 09/Aug/19

thanks

$${thanks} \\ $$

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