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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

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

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

Can you answer this question?
 
 

2 Answers

0 votes
Answer : $\frac{2}{810}\;[10^{n+1}-10-9n]$
Explanation : $\;S_{n}=0.2+0.22+0.222+0.2222\;-----\;n$
$=\frac{0.2}{9}\;[9+99+999+9999+------+n]$
$=\frac{0.2}{9}\;[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+------+(10^n-1)]$
$=\frac{2}{90}\;[10+10^2+10^3+-----+10^n-(1+1+1+----+n\;terms)]$
$=\frac{2}{90}\;[\frac{10(10^n-1)}{10-1}-n]$
$=\frac{2}{90}\;[\frac{10(10^n-1)}{9}-n]$
$=\frac{2}{90}*\frac{1}{9}\;[10^{n+1}-10-9n]$
$=\frac{2}{180}\;[10^{n+1}-10-9n].$
answered Jan 18, 2014 by yamini.v
 
0 votes
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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