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