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Question Number 93474 by Rio Michael last updated on 13/May/20

Q. Prove by mathematical induction that Σ_(r=1) ^n (4r + 5) = 2n^2 + 7n

$$\mathrm{Q}.\:\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\mathrm{4}{r}\:+\:\mathrm{5}\right)\:=\:\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{7}{n}\: \\ $$

Answered by eswar150933 last updated on 13/May/20

Answered by allizzwell23 last updated on 13/May/20

P_1 : 4(1) + 5 = 9 = 2(1)^2 + 7(1) (true) Let n = k Σ_(r=1) ^n (4r + 5) = 2k^2 + 7k Assume n = k + 1 is true Σ_(r=1) ^n (4r + 5) + (4(r+1) + 5) = 2(k+1)^2 + 7(k+1) To prove the truth of n = k + 1 Σ_(r=1) ^n (4r + 5) + (4(r+1) + 5) = 2k^2 + 7k + (4(k+1) + 5) = 2k^2 + 7k + 4k + 4 + 5 = 2k^2 + 4k + 2 + 7k + 7 = 2(k^2 + 2k + 1) + 7(k + 1) = 2(k+1)^2 + 7(k+1) DONE ! 130520

$$\:\:\boldsymbol{{P}}_{\mathrm{1}} \::\:\mathrm{4}\left(\mathrm{1}\right)\:+\:\mathrm{5}\:=\:\mathrm{9}\:=\:\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{7}\left(\mathrm{1}\right)\:\:\:\:\:\:\left(\boldsymbol{\mathrm{true}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{Let}}\:\boldsymbol{{n}}\:=\:\boldsymbol{{k}} \\ $$$$\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\mathrm{4}{r}\:+\:\mathrm{5}\right)\:=\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{7}{k}\: \\ $$$$\:\:\boldsymbol{\mathrm{Assume}}\:\boldsymbol{{n}}\:=\:\boldsymbol{{k}}\:+\:\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}} \\ $$$$\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\mathrm{4}{r}\:+\:\mathrm{5}\right)\:+\:\left(\mathrm{4}\left({r}+\mathrm{1}\right)\:+\:\mathrm{5}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{7}\left({k}+\mathrm{1}\right)\: \\ $$$$\:\:\boldsymbol{\mathrm{To}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{truth}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{n}}\:=\:\boldsymbol{{k}}\:+\:\mathrm{1} \\ $$$$\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\mathrm{4}{r}\:+\:\mathrm{5}\right)\:+\:\left(\mathrm{4}\left({r}+\mathrm{1}\right)\:+\:\mathrm{5}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{7}{k}\:+\:\left(\mathrm{4}\left({k}+\mathrm{1}\right)\:+\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{7}{k}\:+\:\mathrm{4}{k}\:+\:\mathrm{4}\:+\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{4}{k}\:+\:\mathrm{2}\:+\:\mathrm{7}{k}\:+\:\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({k}^{\mathrm{2}} \:+\:\mathrm{2}{k}\:+\:\mathrm{1}\right)\:+\:\mathrm{7}\left({k}\:+\:\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{7}\left({k}+\mathrm{1}\right)\: \\ $$$$\boldsymbol{\mathrm{DONE}}\:! \\ $$$$\mathrm{130520} \\ $$

Commented by Rio Michael last updated on 13/May/20

thanks y′all

$$\mathrm{thanks}\:\mathrm{y}'\mathrm{all} \\ $$

Commented by allizzwell23 last updated on 13/May/20

you′re welcom

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcom} \\ $$

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