CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    \[0.2+0.22+0.222+...\]to n terms is equal to:

    A)  \[\left( \frac{2}{9} \right)-\left( \frac{2}{81} \right)(1-{{10}^{-n}})\]

    B)         \[n-\left( \frac{1}{9} \right)(1-{{10}^{-n}})\]

    C)         \[\left( \frac{2}{9} \right)\left[ n-\left( \frac{1}{9} \right)(1-{{10}^{-n}}) \right]\]

    D)         \[\left( \frac{2}{9} \right)\]

    E)         \[\left( \frac{2}{9} \right)\left[ n-\left( \frac{1}{9} \right)(1-{{10}^{n}}) \right]\]

    Correct Answer: C

    Solution :

    \[0.2+0.22+0.222+....+\text{ }n\text{ }terms\] \[=2(0.1+0.11+0.111+...+n\text{ }terms)\] \[=2\left( \frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+......+n\text{ }terms \right)\] \[=\frac{2}{9}\left( \frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+....+n\text{ }terms \right)\] \[=\frac{2}{9}\left( 1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+.... \right.\]                                                 \[+n\text{ }terms)\] \[=\frac{2}{9}\left[ n-\left( \frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+....n \right) \right]\] \[=\frac{2}{9}\left[ n-\frac{1}{10}\frac{\left[ 1-{{\left( \frac{1}{10} \right)}^{n}} \right]}{\left[ 1-\frac{1}{10} \right]} \right]\] \[=\frac{2}{9}\left[ n-\frac{1}{10}\times \frac{10}{9}.\left[ \frac{{{10}^{n}}-1}{{{10}^{n}}} \right] \right]\] \[=\frac{2}{9}\left[ n-\frac{1}{9}(1-{{10}^{-n}}) \right]\]


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