Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

Commented by FilupSmith last updated on 25/May/16

$$\mathbf{LHS}=S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}$$$$S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}$$$$2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}$$$$2S+\frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}+\frac{1}{2^n}$$$$2S+\frac{1}{2^n}=1+S$$$$\therefore S=1-\frac{1}{2^n}$$$$\therefore \frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=1-\frac{1}{2^n}◼$$

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com