Question

The sum of the series $6+66+666+.......$upto $n$ terms is

• A. $\frac{{10}^{n-1}-9n+10}{81}$
• B. $\frac{2\left({10}^{n+1}-9n-10\right)}{27}$
• C. $\frac{2\left({10}^n-9n-10\right)}{27}$
• D. none of these

Answer 1

The correct option is B

$\frac{2\left({10}^{n+1}-9n-10\right)}{27}$

Explanation for correct option

Let $S=6+66+666+.......$

$$\begin{gathered} \Rightarrow S=6+66+666+....... \\ \Rightarrow S=6\left[1+11+111+......\right] \end{gathered}$$

Multiply $S$by $\frac{1}{9}$

$$\begin{gathered} \therefore S=\frac{6}{9}\left[9+99+999+.......\right] \\ \Rightarrow S=\frac{6}{9}\left[\left(10-1+{10}^2-1+{10}^3-1+............+{10}^n-1\right)\right] \end{gathered}$$

$10,{10}^2,{10}^3,............,{10}^n$ has a geometric progression with first term $a=10$ and $r=10$

sum of geometric progression$=a\left[\frac{r^{n-1}}{r-1}\right]$

$\begin{array}{rcl}\therefore \left(10+{10}^2+{10}^3+............+{10}^n\right)& =& 10\left[\frac{{10}^n-1}{10-1}\right]\\ & =& \frac{10}{9}\left[{10}^n-1\right]\end{array}$

$$\begin{gathered} \therefore 6\left[n+\left(10+{10}^2+{10}^3+............+{10}^n\right)\right]=\frac{6}{9}\left[n+\frac{10}{9}\left({10}^n-1\right)\right] \\ \Rightarrow S=\frac{6}{9}\left[\frac{9n+{10}^{n+1}-10}{9}\right] \\ \Rightarrow S=2\left[\frac{{10}^{n+1}+9n-10}{27}\right] \end{gathered}$$

Hence $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.