Question

What is the sum of the series $0.5+0.55+0.555+....$ to $n$ terms?

• A. $\frac{5}{9}[n-\frac{2}{9}(1-\frac{1}{{10}^n})]$
• B. $\frac{1}{9}[5-\frac{2}{9}(1-\frac{1}{{10}^n})]$
• C. $\frac{1}{9}[n-\frac{5}{9}(1-\frac{1}{{10}^n})]$
• D. $\frac{5}{9}[n-\frac{1}{9}(1-\frac{1}{{10}^n})]$

Answer 1

The correct option is D $\frac{5}{9}[n-\frac{1}{9}(1-\frac{1}{{10}^n})]$

$0.5+0.55+0.555+............nterms$

$=5\{0.1+0.11+0.111+........\}$

$=\frac{5}{9}(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+..........)$

$=\frac{5}{9}(1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+.......nterms)$

$=\frac{5}{9}(n-(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+.........nterms))$

Now, $\frac{1}{10},\frac{1}{100},\frac{1}{1000}$ ,.... are in GP

$\therefore 0.5+0.55+0.555+............nterms$

$=\frac{5}{9}(n-\left(\frac{\frac{1}{10}(1-{\left(\frac{1}{10}\right)}^n)}{1-\frac{1}{10}}\right))$

$=\frac{5}{9}[n-\frac{1}{9}(1-\frac{1}{{10}^n})]$