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Question Number 5723 by Rasheed Soomro last updated on 25/May/16

(1/2)+(1/4)+(1/8)+....+(1/2^n )=1−(1/2^n )  P   r   o   v   e the above for integral n≥1.

$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+....+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} } \\ $$$$\mathrm{P}\:\:\:\mathrm{r}\:\:\:\mathrm{o}\:\:\:\mathrm{v}\:\:\:\mathrm{e}\:\mathrm{the}\:\mathrm{above}\:\mathrm{for}\:\mathrm{integral}\:\mathrm{n}\geqslant\mathrm{1}. \\ $$

Commented by FilupSmith last updated on 25/May/16

LHS=S=(1/2)+(1/4)+(1/8)+...+(1/2^n )  S=(1/2)+(1/4)+...+(1/2^n )  2S=1+(1/2)+(1/4)+...+(1/2^(n−1) )  2S+(1/2^n )=1+(1/2)+(1/4)+...+(1/2^(n−1) )+(1/2^n )  2S+(1/2^n )=1+S  ∴ S=1−(1/2^n )  ∴ (1/2)+(1/4)+...+(1/2^n )=1−(1/2^n )            ■

$$\boldsymbol{\mathrm{LHS}}={S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{2}{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\mathrm{2}{S}+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{2}{S}+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }=\mathrm{1}+{S} \\ $$$$\therefore\:{S}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$

Commented by Rasheed Soomro last updated on 25/May/16

GooD Approach!

$$\mathcal{G}{oo}\mathcal{D}\:\mathcal{A}{pproach}! \\ $$

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