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 137376 by benjo_mathlover last updated on 02/Apr/21

$$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{9}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{45}+\frac{1}{75}+...=?$$

Answered by EDWIN88 last updated on 02/Apr/21

$$\overline{)\begin{array}{cccccccc}& \frac{1}{3^0}& \frac{1}{3^1}& \frac{1}{3^2}& \frac{1}{3^3}& ...& \frac{1}{3^{\mathrm{n}}}& \underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\\ \frac{1}{5^0}& \frac{1}{5^0.3^0}& \frac{1}{5^0.3^1}& \frac{1}{5^0.3^2}& \frac{1}{5^0.3^3}& ...& \frac{1}{5^0.3^{\mathrm{n}}}& \frac{1}{5^0}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{n}}}\\ \frac{1}{5^1}& \frac{1}{5^1.3^0}& \frac{1}{5^1.3^1}& \frac{1}{5^1.3^2}& \frac{1}{5^1.3^3}& ...& \frac{1}{5^1.3^{\mathrm{n}}}& \frac{1}{5^1}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{n}}}\\ ...& ...& ...& ...& ...& ...& ...& ...\\ \frac{1}{5^{\mathrm{m}}}& \frac{1}{5^{\mathrm{m}}.3^0}& \frac{1}{5^{\mathrm{m}}.3^1}& \frac{1}{5^{\mathrm{m}}.3^2}& \frac{1}{5^{\mathrm{m}}.3^3}& ...& \frac{1}{5^{\mathrm{m}}.3^{\mathrm{n}}}& \frac{1}{5^{\mathrm{m}}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{n}}}\end{array}}$$$$\mathrm{so}\mathrm{we}\mathrm{get}:(\frac{1}{5^0}+\frac{1}{5^1}+\frac{1}{5^2}+...+\frac{1}{5^{\mathrm{m}}}).\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{n}}}=$$$$\left(\underset{\mathrm{m}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{5^{\mathrm{m}}}\right).\left(\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{n}}}\right)=\frac{1}{1-\frac{1}{5}}.\frac{1}{1-\frac{1}{3}}=\frac{15}{8}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com