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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

The sum of 0.2, 0.22, 0.222, 0.2222.....till n terms is given by

$(a)\;\frac{2}{9}\;[10^{n+1}-10-9n]\qquad(b)\;\frac{2}{81}\;[10^{n+1}+10+9n]\qquad(c)\;\frac{2}{810}\;[10^{n+1}-10-9n]\qquad(d)\;\frac{2}{81}\;[10^{n+1}-10-9n]$

This is a wrong answer this can also come in chance if u now the than give if u don't now why giving such rubbish answers ok if u don't now then don't give such foolish answer

2 Answers

Explanation : $\;S_{n}=0.2+0.22+0.222+0.2222\;-----\;n\;terms $
$= 2 (0.1 + 0.11+ 0.111+ ------ n\; terms )$
$=\frac{2}{9}\;[0.9 + 0.99 + 0.999 + ------+n\; terms]$
$= \frac{2}{9} [1- 0.1 + 1 - 0.01 + 1- 0.001 + ------+ n\; terms ]$
$= \frac{2}{9}[1 + 1 + 1 + ---+ n \; terms] - [0.1 + 0.01+ 0.001 + ---+ n\; terms]$
$ \begin{align*}=\frac{2}{9} \begin{pmatrix}n - \frac{(1 - (0.1)^n)}{1-0.1} \end{pmatrix}\end{align*}$
$\begin{align*}= \frac{2}{9} \begin{pmatrix}n - \frac{1 - (0.1)^n}{9}\end{pmatrix} \end{align*}$
$= \frac{2}{81}[9n - 1+ (0.1)^n]$
answered Jan 18, 2014 by yamini.v
edited Nov 3 by priyanka.c
 
Sn = 0.2 + 0.22 + 0.222 + ……… to n terms)
Sn = 2(0.1 + 0.11 + 0.111 + ..... to n terms)
Sn = 2 / 9 (0.9 + 0.99 + 0.999 + ..... to n terms)
Sn = 2 / 9 (  1 - 0.1  +   1 - 0.01   +    1 - 0.001   + ..... to n terms)
Sn = 2 / 9 (  1 + 1 + 1 + ... to n terms  - 0.1 - 0.01 - 0.001... to n terms)
AP is 1 + 1 + 1 + ... to n terms
GP is 0.1 + 0.01 + 0.001... 
which will be a(1-r^n)/(1-r)
 
Sn = 2 / 9 (  (1 + 1 + ... to n terms)  - (0.1 + 0.01 + 0.001... to n terms))
Sn = 2 / 9 (  n  - 0.1 (1-0.1^n)/(1-0.1))
Sn = 2 / 9 (  n  - (1-0.1^n)/9)
Sn = 2 / 81 (  9n  - (1-0.1^n))
Sn = 2 / 81 (  9n  - 1 + 0.1^n)
 
answered Sep 16, 2015 by illusionist.nectar
 

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