Answer:
Option C
Explanation:
$k.10^{9}=10^{9}+2(11)^{1}(10)^{8}+3((11)^{2}(10)^{7}+....+10(11)^{9}$
$k=1+2\left(\frac{11}{10}\right)+3\left(\frac{11}{10}\right)^{2}+...+10\left(\frac{11}{10}\right)^{9}$ .......(i)
$\left(\frac{11}{10}\right)k=1\left(\frac{11}{10}\right)+2\left(\frac{11}{10}\right)^{2}+...+9\left(\frac{11}{10}\right)^{9}+10\left(\frac{11}{10}\right)^{10}$ ..........(ii)
On subtracting Eq,(ii) from Eq.(i) , we get
$k\left(1-\frac{11}{10}\right)=1+\left(\frac{11}{10}\right)+\left(\frac{11}{10}\right)^{2}+...+\left(\frac{11}{10}\right)^{9}-10\left(\frac{11}{10}\right)^{10}$
$\Rightarrow$ $ k\left(\frac{10-11}{10}\right)=\frac{1\left[\left(\frac{11}{10}\right)^{10}-1\right]}{\left(\frac{11}{10}-1\right)}-10\left(\frac{11}{10}\right)^{10}$
[ $\because$ In GP sum of n terms = $\frac{a(r^{n}-1)}{r-1}$ when r>1]
$\Rightarrow$ -k
.$=10\left[10\left(\frac{11}{10}\right)^{10}-10-10 \left(\frac{11}{10}\right)^{10}\right]$
k=100
