Find sum of series upto n term of 0.6+0.66+0.666+0.6666

1.	Find sum of series upto n term of 0.6+0.66+0.666+0.6666
Answer» ANSWER: $S_n=\frac{6}{9}[n-\frac{1}{9}(1-\frac{1}{10^n})]$ Step-by-step explanation: GIVEN : Series 0.6+0.66+0.666+0.6666 To find :The SUM of series up-to n terms Solution : Let $S_n$ denotes the sum of n terms. $S_n=0.6+0.66+0.666+0.6666 ........+\text{n terms}$ $S_n=6[0.1+0.11+0.111+0.1111 ........+\text{n terms}]$ $S_n=\frac{6}{9}[0.9+0.99+0.999+ ........+\text{n terms}]$ $S_n=\frac{6}{9}[(1-\frac{1}{10})+(1-\frac{1}{10^2})+(1-\frac{1}{10^3})+ ........+(1-\frac{1}{10^n})]$ $S_n=\frac{6}{9}[(1+1+1....+1)-(\frac{1}{10})+\frac{1}{10^2}+\frac{1}{10^3}+ ........+\frac{1}{10^n})]$ Sum of (1+1+1+1.....1) n TIMES = n Apply G.P series in second BRACKET in which $a=\frac{1}{10} , r=\frac{1}{10}$ Sum of G.P series is $S_n=\frac{a(r^n-1)}{r-1}$ $S_n=\frac{6}{9}[(n)-(\frac{\frac{1}{10}((\frac{1}{10})^n-1)}{\frac{1}{10}-1})]$ $S_n=\frac{6}{9}[(n)-\frac{1}{9}(1-\frac{1}{10^n})]$ Therefore, The sum of series of n terms is given by, $S_n=\frac{6}{9}[n-\frac{1}{9}(1-\frac{1}{10^n})]$

Answer»

$S_n=\frac{6}{9}[n-\frac{1}{9}(1-\frac{1}{10^n})]$