Question

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

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

Answer 1

Answer: (B)

Let, $S_n=9+99+999+\dots \dots n$ terms
$\Rightarrow S_n=(10-1)+(100-1)+(1000-1)+\dots n$ terms
$\Rightarrow S_n=(10+10^2+10^3+\dots \dots n$ terms $)$
$-(1+1+\dots \dots n$ terms $)$
$\Rightarrow S_n=\frac{10\left(10^n-1\right)}{10-1}-n$
$\left[\because a+ar+a{r}^2+\dots \dots +a{r}^{n-1}=\frac{a\left(r^n-1\right)}{r-1},r>1\right]$
$\Rightarrow S_n=\frac{10}{9}\left(10^n-1\right)-n$
Put $n=10$
$\Rightarrow S_{10}=\frac{10}{9}\left(10^{10}-1\right)-10$
$=\frac{10}{9}\left(10^{10}-1-9\right)$
$=\frac{10}{9}\left(10^{10}-10\right)=\frac{100}{9}\left(10^9-1\right)$