Question

The sum of the first 10 terms of the series $9+99+999+.....$ is

• A. $\frac{9}{8}(9^{10}-1)$
• B. $\frac{100}{9}({10}^9-1)$
• C. ${10}^9-1$
• D. $\frac{100}{9}({10}^{10}-1)$

Answer 1

Answer: (B)

$\frac{100}{9}({10}^9-1)$

$9+99+999+.................10terms\Rightarrow (10-1)+(100-1)+...........10terms\Rightarrow (10+100+1000+.............10terms)-(1+1+1+......10times)\Rightarrow \frac{10[{10}^{10}-1]}{10-1}-10\Rightarrow \frac{{10}^{11}-10-90}{9}=\frac{{10}^{11}-100}{9}=\frac{100({10}^9-1)}{9}$
Therefore Answer is $B$