How do you prove that10n−110n−2=1+∑i=0∞2i×10−n(i+1)

Bradlee Hooper

Answered question

2022-02-27

How do you prove that $\frac{{10}^{n}-1}{{10}^{n}-2}=1+\sum _{i=0}^{\mathrm{\infty}}{2}^{i}\times {10}^{-n(i+1)}$

Answer & Explanation

Elijah Hunt

Beginner2022-02-28Added 5 answers

Your sum is a geometric series: $\sum _{i=0}^{\mathrm{\infty}}{2}^{i}\times {10}^{-n(i+1)}={10}^{-n}\sum _{i=0}^{\mathrm{\infty}}{\left(\frac{2}{{10}^{n}}\right)}^{i}$ $={10}^{-n}\times \frac{1}{1-2\times {10}^{-n}}=\frac{1}{{10}^{n}-2}$ You are done once you observe that $\frac{{10}^{n}-1}{{10}^{n}-2}=\frac{{10}^{n}-2}{{10}^{n}-2}+\frac{1}{{10}^{n}-2}$ $=1+\frac{1}{{10}^{n}-2}$