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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Sequences and Series

NDA 2015 · Mathematics · Sequences and Series

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

What is the sum of the series 0.5+0.55+0.555+…0.5+0.55+0.555+\ldots to nn terms?

Options

  1. A59\lleft[n−29\l\frac{5}{9} \l left[n-\frac{2}{9}\l lleft(1-110n)]\frac{1}{10^n})]
  2. B19\lleft[5−29\l\frac{1}{9} \l left[5-\frac{2}{9}\l lleft(1-110n)]\frac{1}{10^n})]
  3. C19\lleft[n−59\l\frac{1}{9} \l left[n-\frac{5}{9}\l lleft(1-110n)]\frac{1}{10^n})]
  4. D59\lleft[n−19\l\frac{5}{9} \l left[n-\frac{1}{9}\l lleft(1-110n)]\frac{1}{10^n})]

Correct answer

D. 59\lleft[n−19\l\frac{5}{9} \l left[n-\frac{1}{9}\l lleft(1-110n)]\frac{1}{10^n})]

Step-by-step solution

Given 0.5 + 0.55 + 0.555 + ..... to n = 5 [0.1 + 0.11 + 0.111 + ..... to n terms] =59[0.9+0.99+0.999+….. to n terms ]=59[910+99100+9991000+…. 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}=59((1−110)+(1−1102)+(1−1103)+…⋅.(1−110n)]=59[n−(110+1102+…⋅110n)]\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}\l$lleft(1-\frac{1}{10^{\mathrm{n}}}\right)]
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