logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

Find the sum of series 4 + 44 + 444 + 4444 +------upto n terms.

$\begin{array}{1 1} \infty \\ \frac{4}{9}[\frac{10(10^n-1)}{9}-n] \\44^4-4 \\ \frac{4(4^n-1)}{3}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (b) $\frac{4}{9}[\frac{10(10^n-1)}{9}-n]$
Explanation: S=4+44+444+-------n terms
$=4(1+11+111+1111+------n\;terms)$
$=\frac{4}{9}(9+99+999+-----------n\;terms)$
$=\frac{4}{9}(10-1+100-1+1000-1+-----------n\;terms)$
$=\frac{4}{9}((10-1)+(10^2-1)+(10^3-1)+-----------n\;terms)$
$=\frac{4}{9}[10+10^2+10^3+-----------n\;terms-n]$
$=\frac{4}{9}[\frac{10(10^n-1)}{10-1}-n]$
$=\frac{4}{9}[\frac{10(10^n-1)}{9}-n].$
answered Dec 31, 2013 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...