Question

Let $P=\frac{3}{17}+\frac{33}{{17}^2}+\frac{333}{{17}^3}+\cdots \infty$ then $P$ equals

• A. $\frac{3}{17}$
• B. $\frac{7}{17}$
• C. $\frac{3}{7}$
• D. $\frac{51}{112}$

Answer 1

The correct option is D $\frac{51}{112}$

$P=\frac{3}{17}+\frac{33}{{17}^2}+\frac{333}{{17}^3}+....\infty$_____(1)
$\frac{P}{17}=\frac{3}{{17}^2}+\frac{33}{{17}^3}+....\infty$_______(2)
$\left(2\right)-\left(1\right)\Rightarrow$
$P(1-\frac{1}{17})=\frac{3}{17}+\frac{30}{{17}^2}+\frac{300}{{17}^3}+....\infty$
$=\frac{3}{17}(1+\frac{10}{17}+\frac{100}{{17}^2}+...\infty$
$=\frac{3}{17}\left(\frac{1}{1-\frac{10}{17}}\right)=\frac{3}{7}$
$\Rightarrow \frac{16P}{17}=\frac{3}{7}$
$\therefore P=\frac{51}{112}$