Question

Sum of series $6+66+666+...$ up to $n$ terms is

• A. $\frac{2}{27}({10}^n-9n-10)$
• B. $\frac{2}{27}({10}^{n+1}-9n-10)$
• C. $\frac{1}{9}({10}^n-9n-10)$
• D. $\frac{1}{9}({10}^{n+1}-9n+10)$

Answer 1

The correct option is B $\frac{2}{27}({10}^{n+1}-9n-10)$

Given $S_n=6+66+666+...$

$\Rightarrow S_n=6(1+11+111+.....)$

$\Rightarrow S_n=6[1+(1+10)+(1+10+{10}^2)+.....]$

Now, $t_r=6(1+10+{10}^2+....+{10}^{r-1})=\frac{6({10}^r-1)}{10-1}$ [ Sum of GP $=\frac{a(n^r-1)}{n-1}$ when $n>1$]

$\Rightarrow t_r=\frac{2({10}^r-1)}{3}$

$S_n={\sum }_{r=1}^n{t}_r=\frac{2}{3}{\sum }_{r=1}^n({10}^r-1)=\frac{2}{3}{\sum }_{r=1}^n{10}^r-{\sum }_{r=1}^n\frac{2}{3}$

$\Rightarrow S_n=\frac{2}{3}\left[\frac{10({10}^n-1)}{10-1}\right]-\frac{2n}{3}=\frac{2}{27}({10}^{n+1}-9n-10)$

Ans: B