Question

Value of $9+99+999+....$ upto $n$ terms is

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

Answer 1

Answer: (D)

$\frac{{10}^{n+1}-9n-10}{9}$

Given, $9+99+999+....$ upto $n$ terms
we can write this series in this form
$s=(10-1)+(100-1)+(1000-1).......$ up to $n$ terms.
$=\left[\right(10+100+1000+......)+(-1-1-1-.......\text{up to n terms}\left)\right]$
$=\left[\right(10+{10}^2+{10}^3+.....\text{up to n terms})-n]$ ........(i)
sum of n terms of GP. $=\frac{a(r^n-1)}{r-1}$, where $a$ is first term and $r$ is the common ratio of the GP.
$\text{We have the GP}10+{10}^2+{10}^3+....\text{up to n terms}$, here first term $a=10$ and common ratio $r=10$
So,$10+{10}^2+{10}^3+......\text{n terms}=\frac{10({10}^n-1)}{10-1}=\frac{{10}^{n+1}-10}{9}$On Substituting$s=10+{10}^2+{10}^3+......\text{n terms}=\frac{{10}^{n+1}-10}{9}$ in equation (i). we get,
$=\frac{{10}^{n+1}-10}{9}-n=\frac{{10}^{n+1}-9n-10}{9}$
$\Rightarrow 9+99+999+.......\text{up to n terms}=\frac{{10}^{n+1}-9n-10}{9}$
Option D is correct.