Question Number 199987 by Yakubu last updated on 11/Nov/23
![Use mathematical induction to prove that the statement a+(a+d)+(a+2d)+...+(a+(n−1)d)=(n/2)[2a+(n−1)d] is true for all natural numbers Any help please](Q199987.png)
$$\:\boldsymbol{{Use}}\:\boldsymbol{{mathematical}}\:\boldsymbol{{induction}} \\ $$$$\:\boldsymbol{{to}}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{statement}} \\ $$$$\:\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{d}}\right)+\left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{d}}\right)+...+\left(\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right)=\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right] \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{natural}}\:\boldsymbol{\mathrm{numbers}} \\ $$$$\boldsymbol{\mathrm{A}{ny}}\:\boldsymbol{{help}}\:\boldsymbol{{please}} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Nov/23
![p(n): a+(a+d)+(a+2d)+...+(a+(n−1)d)=(n/2)[2a+(n−1)d] •p(1): a+(1−1)d=(1/2)(2a+(1−1)d) a=a p(1) is true • let p(k) is true: p(k): a+(a+d)+(a+2d)+...+(a+(k−1)d)=(k/2)[2a+(k−1)d] ....+(a+(k−1)d)+a+(k+1^(−) −1)d=(k/2)[2a+(k−1)d]+a+(k+1−1)d =ka+k(((k−1)/2))d+a+kd =a(k+1)+(((k^2 −k)/2)+k)d =a(k+1)+(((k^2 −k+2k)/2))d =2a(((k+1)/2))+(((k^2 +k)/2))d =2a(((k+1)/2))+k(((k+1)/2))d =2a(((k+1)/2))[2a+kd] =2a(((k+1)/2))[2a+(k+1^(−) −1)d] ∴ p(k+1) is true when p(k) is true Hence p(n) is true ∀n∈N](Q199991.png)
$$\:{p}\left({n}\right):\:\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{d}}\right)+\left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{d}}\right)+...+\left(\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right)=\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right] \\ $$$$\bullet{p}\left(\mathrm{1}\right):\:\mathrm{a}+\left(\mathrm{1}−\mathrm{1}\right)\mathrm{d}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2a}+\left(\mathrm{1}−\mathrm{1}\right)\mathrm{d}\right) \\ $$$$\mathrm{a}=\mathrm{a} \\ $$$$\mathrm{p}\left(\mathrm{1}\right)\:{is}\:{true} \\ $$$$\bullet\:{let}\:{p}\left({k}\right)\:{is}\:{true}: \\ $$$$\:{p}\left({k}\right):\:\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{d}}\right)+\left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{d}}\right)+...+\left(\boldsymbol{\mathrm{a}}+\left({k}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right)=\frac{{k}}{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\mathrm{a}}+\left({k}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right] \\ $$$$\:\:\:\:....+\left(\boldsymbol{\mathrm{a}}+\left({k}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right)+\mathrm{a}+\left(\overline {{k}+\mathrm{1}}−\mathrm{1}\right)\mathrm{d}=\frac{{k}}{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\mathrm{a}}+\left({k}−\mathrm{1}\right)\boldsymbol{\mathrm{d}}\right]+\mathrm{a}+\left({k}+\mathrm{1}−\mathrm{1}\right)\mathrm{d} \\ $$$$\:\:={k}\mathrm{a}+{k}\left(\frac{{k}−\mathrm{1}}{\mathrm{2}}\right)\mathrm{d}+\mathrm{a}+{k}\mathrm{d} \\ $$$$\:\:=\mathrm{a}\left({k}+\mathrm{1}\right)+\left(\frac{{k}^{\mathrm{2}} −{k}}{\mathrm{2}}+{k}\right){d} \\ $$$$\:\:=\mathrm{a}\left({k}+\mathrm{1}\right)+\left(\frac{{k}^{\mathrm{2}} −{k}+\mathrm{2}{k}}{\mathrm{2}}\right){d} \\ $$$$\:\:=\mathrm{2a}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{{k}^{\mathrm{2}} +{k}}{\mathrm{2}}\right){d} \\ $$$$\:\:=\mathrm{2a}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)+{k}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right){d} \\ $$$$\:\:=\mathrm{2a}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)\left[\mathrm{2a}+{kd}\right] \\ $$$$\:\:=\mathrm{2a}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)\left[\mathrm{2a}+\left(\overline {{k}+\mathrm{1}}−\mathrm{1}\right){d}\right] \\ $$$$\therefore\:{p}\left({k}+\mathrm{1}\right)\:{is}\:{true}\:{when}\:{p}\left({k}\right)\:{is}\:{true} \\ $$$${Hence}\:{p}\left({n}\right)\:{is}\:{true}\:\forall{n}\in\mathbb{N} \\ $$
Commented by Yakubu last updated on 12/Nov/23
$${thank}\:{you}\:{sir} \\ $$