Discussion Forum

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

Answer:

Explanation:

Let ,

 S_n= 9+99+999+.........n terms

 $\Rightarrow$    $S_{n}=(10-1)+(100-1)+(1000-1)+$.... n  terms

 $\Rightarrow$    $S_{n}=(10+10^{2}+10^{3}+.......$   n terms -(1+1+..... n terms)

$\Rightarrow$     $S_{n}=\frac{10(10^{n}-1)}{10-1}-n$

$[a+ar+ar^{2}+..... ar^{n-1}= \frac{a(r^{n}-1}{r-1}, r>1]$

 $\Rightarrow$    $S_{n}= \frac{10}{9}(10^{n}-1)-n$

 put n=10

$\Rightarrow$    $S_{10}= \frac{10}{9}(10^{10}-1)-10$

 $= \frac{10}{9}(10^{10}-1-9)$

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