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Question Number 2762 by Rasheed Soomro last updated on 26/Nov/15

Without using induction_(−) or arithmatic series−concept _(−) prove the following: 1+2+3+...+n=((n(n+1))/2)

$${Without}\:{using}\:\underset{−} {{induction}}\:{o}\underset{−} {{r}\:\:{arithmatic}\:{series}−{concept}\:\:\:} \\ $$$$\:{prove}\:{the}\:{following}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$

Answered by prakash jain last updated on 26/Nov/15

n^2 −(n−1)^2 =2n−1 1^2 −0^2 =2∙1−1 2^2 −1^2 =2∙2−1 3^2 −2^2 =3∙2−1 ... ... n^2 −(n−1)^2 =2∙n−1 Sum all of the above. n^2 =2(1+2+..+n)−n 2(1+2+3+..+n)=n^2 +n Σ_(i=1) ^n i=((n(n+1))/2)

$${n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{n}−\mathrm{1} \\ $$$$\mathrm{1}^{\mathrm{2}} −\mathrm{0}^{\mathrm{2}} =\mathrm{2}\centerdot\mathrm{1}−\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\mathrm{2}\centerdot\mathrm{2}−\mathrm{1} \\ $$$$\mathrm{3}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} =\mathrm{3}\centerdot\mathrm{2}−\mathrm{1} \\ $$$$... \\ $$$$... \\ $$$${n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}\centerdot{n}−\mathrm{1} \\ $$$$\mathrm{Sum}\:\mathrm{all}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above}. \\ $$$${n}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\mathrm{2}+..+{n}\right)−{n} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+{n}\right)={n}^{\mathrm{2}} +{n} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$

Commented by Rasheed Soomro last updated on 27/Nov/15

ThankS!

$$\mathcal{T}{hank}\mathcal{S}! \\ $$

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