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Question Number 2759 by RasheedAhmad last updated on 26/Nov/15

Without using arithmatic Series concept or formula prove the following: 1+2+3+...+n=((n(n+1))/2)

$${Without}\:{using}\:{arithmatic}\:{Series} \\ $$$${concept}\:{or}\:{formula}\:{prove}\:{the}\:{following}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$

Answered by 123456 last updated on 26/Nov/15

induction 1+2+∙∙∙+n=((n(n+1))/2) base case: n=1 ((1(1+1))/2)=(2/2)=1 induction step: suppose its true for n, lets shown its true for n+1 1+2+∙∙∙+n+n+1=((n(n+1))/2)+n+1 =(n+1)((n/2)+1) =(((n+1)(n+2))/2) =(((n+1)[(n+1)+1])/2) so, since its true for n=1, then its true to n=1,2,3,4,.... n∈N^∗

$$\mathrm{induction} \\ $$$$\mathrm{1}+\mathrm{2}+\centerdot\centerdot\centerdot+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{base}\:\mathrm{case}:\:{n}=\mathrm{1} \\ $$$$\frac{\mathrm{1}\left(\mathrm{1}+\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{induction}\:\mathrm{step}: \\ $$$$\mathrm{suppose}\:\mathrm{its}\:\mathrm{true}\:\mathrm{for}\:{n},\:\mathrm{lets}\:\mathrm{shown}\:\mathrm{its}\:\mathrm{true} \\ $$$$\mathrm{for}\:{n}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2}+\centerdot\centerdot\centerdot+{n}+{n}+\mathrm{1}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n}+\mathrm{1} \\ $$$$=\left({n}+\mathrm{1}\right)\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\left[\left({n}+\mathrm{1}\right)+\mathrm{1}\right]}{\mathrm{2}} \\ $$$$\mathrm{so},\:\mathrm{since}\:\mathrm{its}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{1},\:\mathrm{then}\:\mathrm{its}\:\mathrm{true} \\ $$$$\mathrm{to}\:{n}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},.... \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$

Commented by Rasheed Soomro last updated on 26/Nov/15

Nice! Actually at first I had also included the condition ′without induction′ but in editing the question this has been deleted. Now I have posted the question again with the mentioned condition.

$$\mathcal{N}{ice}! \\ $$$${Actually}\:{at}\:{first}\:{I}\:{had}\:{also}\:\:{included}\:{the}\:{condition} \\ $$$$'{without}\:{induction}'\:{but}\:{in}\:{editing}\:{the}\:{question}\:{this} \\ $$$${has}\:{been}\:{deleted}.\:{Now}\:{I}\:{have}\:{posted}\:{the}\:{question} \\ $$$${again}\:{with}\:{the}\:{mentioned}\:\:{condition}. \\ $$

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