NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### WB JEE 2009

Subjective

Find the sum of the first n terms of the series 0.2 + 0.22 + 0.222 + ......

.

## Explanation

Sn = 0.2 + 0.22 + 0.222 + ..... n terms

$${S_n} = {2 \over 9}(0.9 + 0.99 + 0.999 + \,\,.......\,n\,terms)$$

$${S_n} = {2 \over 9}\left( {{{10 - 1} \over {10}} + {{100 - 1} \over {100}} + {{1000 - 1} \over {1000}} + \,\,....\,n\,terms} \right)$$

$$= {2 \over 9}\left( {1 - {1 \over {10}} + 1 - {1 \over {{{(10)}^2}}} + 1 - {1 \over {{{(10)}^3}}} + \,\,....\,n\,terms} \right)$$

$$= {2 \over 9}\left( {n - \left( {{1 \over {10}} + {1 \over {{{(10)}^2}}} + \,\,....\,n\,terms} \right)} \right)$$

$$= {2 \over 9}\left( {n - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \over {1 - {1 \over {10}}}}} \right) = {2 \over 9}\left( {n - {1 \over {10}} \times {{10} \over 9}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \right)$$

$$= {2 \over 9}\left( {n - {1 \over 9}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \right) = {{2n} \over 9} - {2 \over {81}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12