Question

8 + 88 + 888 + .............n terms = ____

• A. $\frac{80}{81}({10}^n-1)-\frac{8n}{9}$
• B. $\frac{10}{81}({10}^n-1)$
• C. $\frac{80}{81}({10}^n-1)+\frac{8n}{9}$
• D. $\frac{80}{81}({10}^n-1)-\frac{n}{9}$

Answer 1

The correct option is A

$\frac{80}{81}({10}^n-1)-\frac{8n}{9}$

$S=8+88+888+.....=8[1+11+\mathrm{111....}]=\frac{8}{9}[9+99+999+......]=\frac{8}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+.....]=\frac{8}{9}\left[\right(10+{10}^2+{10}^3+......)-(1+1+1+......\left)\right]=\frac{8}{9}[10\times \frac{{10}^n-1}{10-1}-n]$
$=\frac{8}{9}[10\times \frac{{10}^n-1}{9}-n]$
$=\frac{80}{81}({10}^n-1)-\frac{8n}{9}$