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Question Number 128112 by AgnibhoMukhopadhyay last updated on 04/Jan/21

$$1+2+3+4+.....+100=?$$

Answered by Olaf last updated on 04/Jan/21

$$\mathrm{A}(n+1)\times (n+1)\mathrm{squared}\mathrm{chess}\mathrm{board}$$$$\mathrm{contains}{(n+1)}^2\mathrm{squares}.$$$$$$$$\mathrm{and}:$$$$$$$$-\mathrm{Its}\mathrm{diagonal}\mathrm{contains}n+1\mathrm{squares}$$$$-\mathrm{Its}\mathrm{upper}\mathrm{part}\mathrm{contains}n+(n-1)+$$$$(n-2)+...+3+2+1={\mathrm{S}}_n\mathrm{squares}$$$$-\mathrm{Its}\mathrm{lower}\mathrm{part}\mathrm{contains}1+2+3+...+$$$$(n-2)+(n-1)+n={\mathrm{S}}_n\mathrm{squares}$$$$$$$$\Rightarrow {(n+1)}^2=(n+1)+{2\mathrm{S}}_n$$$$\Rightarrow {\mathrm{S}}_n=\frac{{(n+1)}^2-(n+1)}{2}=\frac{n(n+1)}{2}$$

Answered by Ar Brandon last updated on 04/Jan/21

$${\mathrm{S}}_{100}=\frac{100}{2}(1+100)=50\times 101=5050$$

Answered by Geovanek last updated on 04/Jan/21

$$1+2+3+4+.....+100=x$$$$100+99+98+97+.....+1=x$$$$101+101+101+101+.....+101=2x$$$$101\cdot 100=2x$$$$\frac{101\cdot 100}{2}=x$$$$101\cdot 50=x$$$$5050=x$$$$\mathrm{So},\mathrm{we}\mathrm{have}$$$$1+2+3+4+.....+100=5050$$

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