Question

The area of regular polygon of n sides with length of side $\sqrt{3}$, where n > 1 and satisfies the relation $\sum _{r=0}^n\frac{n^2-3n+3}{2{}^n{C}_r}=\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ is :

• A. $30\sqrt{3}$
• B. $\frac{3\sqrt{3}}{4}$
• C. $\frac{5\sqrt{3}}{4}$
• D. $\frac{3\sqrt{3}}{2}$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}$

$\sum \frac{r}{{}^n{C}_r}=\sum \frac{n-r}{{}^n{C}_r}=\frac{n}{2}\sum \frac{1}{{}^n{C}_r}$

$\frac{n^2-3n+3}{2}\sum \frac{1}{{}^n{C}_r}\Rightarrow n^2-3n+3=n$

$n^2-4n+3=0\Rightarrow n=3(\because n>1)$

$\therefore$ Area of the equilateral triangle with side $\sqrt{3}$ is $\frac{3\sqrt{3}}{4}$