Question

If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+...+10(11{)}^9=k(10{)}^9$, then $k$ is equal to :

• A. $\frac{121}{10}$
• B. $\frac{441}{100}$
• C. $100$
• D. $110$

Answer 1

The correct option is C $100$

$(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+...+10(11{)}^9=k(10{)}^9$
$\Rightarrow k=1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+...+10{\left(\frac{11}{10}\right)}^9...\left(1\right)$
Multiply equation (1) with $\frac{11}{10}$
$\Rightarrow \frac{11}{10}k=\frac{11}{10}+2{\left(\frac{11}{10}\right)}^2+3{\left(\frac{11}{10}\right)}^3+...+9{\left(\frac{11}{10}\right)}^9+10{\left(\frac{11}{10}\right)}^{10}$
Substracting equation (2) from equation (1)
$\Rightarrow -\frac{k}{10}=1+\frac{11}{10}+{\left(\frac{11}{10}\right)}^2+{\left(\frac{11}{10}\right)}^3+...+{\left(\frac{11}{10}\right)}^9-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow -\frac{k}{10}=\frac{1[{\left(\frac{11}{10}\right)}^{10}-1]}{\frac{11}{10}-1}-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow k=100$