Question

Find the sum of $n$ terms of the series.
$0.5+0.55+0.555+....n$ terms.

Answer 1

Given series $0.5+0.55+\mathrm{0.555...}n$ terms

we know that,

$0.1+{0.1}^2+{0.1}^3+....=\frac{0.1(1-{0.1}^n)}{0.9}=\frac{(1-{0.1}^n)}{9}\Rightarrow 5(0.1+0.11+0.111+..)\Rightarrow 5(\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+...)\Rightarrow \frac{5}{9}(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+..)\Rightarrow \frac{5}{9}((1-\frac{1}{10})+(1-\frac{1}{100})+(1-\frac{1}{1000})+..)\Rightarrow \frac{5}{9}((1+1+...\text{n terms})-(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...))\Rightarrow \frac{5}{9}(n-\frac{(1-{0.1}^n)}{9})$