Q177 / 270 · NDA 2015 (Phase 1) · Mathematics
What is the sum of the series $0.5+0.55+0.555+\ldots$ to $n$ terms?
- A$\frac{5}{9}\left[n-\frac{2}{9}\left(1-\frac{1}{10^n}\right)\right]$
- B$\frac{1}{9}\left[5-\frac{2}{9}\left(1-\frac{1}{10^n}\right)\right]$
- C$\frac{1}{9}\left[n-\frac{5}{9}\left(1-\frac{1}{10^n}\right)\right]$
- Correct answerD$\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]$
Solution
Given 0.5 + 0.55 + 0.555 + ..... to n
= 5 [0.1 + 0.11 + 0.111 + ..... to n terms]
$\begin{aligned}
& =\frac{5}{9}[0.9+0.99+0.999+\ldots . . \text { to } \mathrm{n} \text { terms }] \\
& =\frac{5}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots . \text { to } \mathrm{n} \text { terms }\right]
\end{aligned}$
$\begin{aligned} & =\frac{5}{9}\left(\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right)+\ldots \cdot .\right. \\ & \left.\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right] \\ & \quad=\frac{5}{9}\left[\mathrm{n}-\left(\frac{1}{10}+\frac{1}{10^2}+\ldots \cdot \frac{1}{10^{\mathrm{n}}}\right)\right]\end{aligned}$
$=\frac{5}{9}\left[\mathrm{n}-\frac{1}{10} \frac{\left\{1-\left(\frac{1}{10}\right)^{\mathrm{n}}\right\}}{\left(1-\frac{1}{10}\right)}\right]=\frac{5}{9}\left[\mathrm{n}-\frac{1}{9}\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right]$
Hence, the correct answer is option: 'D'.