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Question Number 146102 by tabata last updated on 10/Jul/21

prove by mathmatical indiction 5+7+9+.....+(4n+1)=2n^2 +3n

$${prove}\:{by}\:{mathmatical}\:{indiction}\: \\ $$$$\mathrm{5}+\mathrm{7}+\mathrm{9}+.....+\left(\mathrm{4}{n}+\mathrm{1}\right)=\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n} \\ $$

Answered by gsk2684 last updated on 11/Jul/21

let S(n):5+7+9+...+(4n+1)=2n^2 +3n basic step:if n=1 then 5=2(1)^2 +3(1)is true ∴ S(1) is true assumption step: assume S(k) is true 5+7+9+...+(4k+1)=2k^2 +3k inductive step: to prove S(k+1) is true 5+7+9+...+(4k+1)+(4(k+1)+1) =2k^2 +3k+(4k+4+1) =2k^2 +7k+5 =2(k^2 +2k+1)+3(k+1) =2(k+1)^2 +3(k+1) ∴S(k+1) is true ∴By principle of finite mathematical induction given ststement S(n) is true ∀ n∈N

$${let}\:{S}\left({n}\right):\mathrm{5}+\mathrm{7}+\mathrm{9}+...+\left(\mathrm{4}{n}+\mathrm{1}\right)=\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n} \\ $$$$\boldsymbol{{basic}}\:\boldsymbol{{step}}:{if}\:{n}=\mathrm{1}\:{then}\:\mathrm{5}=\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}\right){is}\:{true} \\ $$$$\therefore\:{S}\left(\mathrm{1}\right)\:{is}\:{true}\: \\ $$$$\boldsymbol{{assumption}}\:\boldsymbol{{step}}:\: \\ $$$${assume}\:{S}\left({k}\right)\:{is}\:{true}\: \\ $$$$\mathrm{5}+\mathrm{7}+\mathrm{9}+...+\left(\mathrm{4}{k}+\mathrm{1}\right)=\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k} \\ $$$$\boldsymbol{{inductive}}\:\boldsymbol{{step}}:\: \\ $$$${to}\:{prove}\:{S}\left({k}+\mathrm{1}\right)\:{is}\:{true}\: \\ $$$$\mathrm{5}+\mathrm{7}+\mathrm{9}+...+\left(\mathrm{4}{k}+\mathrm{1}\right)+\left(\mathrm{4}\left({k}+\mathrm{1}\right)+\mathrm{1}\right) \\ $$$$=\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}+\left(\mathrm{4}{k}+\mathrm{4}+\mathrm{1}\right) \\ $$$$=\mathrm{2}{k}^{\mathrm{2}} +\mathrm{7}{k}+\mathrm{5} \\ $$$$=\mathrm{2}\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({k}+\mathrm{1}\right) \\ $$$$\therefore{S}\left({k}+\mathrm{1}\right)\:{is}\:{true}\: \\ $$$$\therefore{By}\:{principle}\:{of}\:{finite}\:{mathematical} \\ $$$${induction}\:{given}\:{ststement}\: \\ $$$${S}\left({n}\right)\:{is}\:{true}\:\forall\:{n}\in{N} \\ $$

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