Question

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

• A. $99$
• B. $100$
• C. $101$
• D. $110$

Answer 1

The correct option is B $100$

$(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\cdots +10(11{)}^9=k(10{)}^9$
Dividing by $(10{)}^9$ on both sides, we get
$1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+\cdots +10{\left(\frac{11}{10}\right)}^9=k\cdots \left(1\right)$
Multiplying eqn $\left(1\right)$ by $\frac{11}{10}$, we get
$\frac{11}{10}+2{\left(\frac{11}{10}\right)}^2+3{\left(\frac{11}{10}\right)}^3+\cdots +10{\left(\frac{11}{10}\right)}^{10}=\left(\frac{11}{10}\right)k\cdots \left(2\right)$

Subtracting eqn $\left(2\right)$ from eqn $\left(1\right)$, we get
$k-\left(\frac{11}{10}\right)k=1+\frac{11}{10}+{\left(\frac{11}{10}\right)}^2+{\left(\frac{11}{10}\right)}^3+\cdots +{\left(\frac{11}{10}\right)}^9-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow k(1-\frac{11}{10})=\frac{{\left(\frac{11}{10}\right)}^{10}-1}{\frac{11}{10}-1}-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow \frac{-k}{10}=10{\left(\frac{11}{10}\right)}^{10}-10-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow k=100$