The question is about sum of a series. A series is given and we need to find out the sum of the series. With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.

Question 18: Find the sum of the series .4 + .44 + .444……. to n terms

- 5.69
- 14.44
- \\frac{4}{81}\\)[9n-1+\\frac{1}{10^n}\\)]
- \\frac{4}{81}\\)[n + 1]

\\frac{4}{81}\\)[9n-1+\\frac{1}{10^n}\\)]

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.4 + .44 + .444+……. to n terms

= 4 * [0.1 +0.11 +0.111+……..to n terms]

= \\frac{4}{9}\\) * [0.9 +0.99 + 0.999+…….to n terms] (Multiplying and dividing by 9)

= \\frac{4}{9}\\) * [(1-0.1)+(1-0.01)+(1-0.001)+…….to n terms]

= \\frac{4}{9}\\) * [(1+1+1…..to n terms)-(0.1+0.01+0.001…to n terms)]

= \\frac{4}{9}\\) * [ n - \\frac{\frac{1}{10} * (1 - \frac{1}{10^n})}{1 - \frac{1}{10}}\\) ]

= \\frac{4}{9}\\) * [n - \\frac{10^n - 1}{9 * 10^n}\\) ]

= \\frac{4}{9}\\) * [n - \\frac{1}{9}\\) * (1 - \\frac{1}{10^n}\\))]

= \\frac{4}{81}\\) * [9n - (1 - \\frac{1}{10^n}\\)]

= \\frac{4}{81}\\) * [9n - 1 + \\frac{1}{10^n}\\)]

The question is **"Find the sum of the series .4 + .44 + .444……. to n terms"**

Choice C is the correct answer.

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