Question

The value of ${\cot }^{-1}\frac{5}{\sqrt{3}}+{\cot }^{-1}\frac{9}{\sqrt{3}}+{\cot }^{-1}\frac{15}{\sqrt{3}}+{\cot }^{-1}\frac{23}{\sqrt{3}}+\cdots \infty$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is B $\frac{\pi }{3}$

Let $S=5+9+15+23+\cdots n\text{terms}$
$S=5+9+15+\cdots$
Subtracting the above two equations,
$t_n=5+(4+6+8+\cdots (n-1\left)\text{terms}\right)$
$\Rightarrow t_n=5+\frac{n-1}{2}[2\times 4+(n-2\left)2\right]$
$\Rightarrow t_n=n^2+n+3$

${\cot }^{-1}\frac{5}{\sqrt{3}}+{\cot }^{-1}\frac{9}{\sqrt{3}}+{\cot }^{-1}\frac{15}{\sqrt{3}}+{\cot }^{-1}\frac{23}{\sqrt{3}}+\cdots \infty$
$=\sum _{r=1}^{\infty }{\cot }^{-1}\left(\frac{r^2+r+3}{\sqrt{3}}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{\sqrt{3}}{r(r+1)+3}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{\frac{r+1}{\sqrt{3}}-\frac{r}{\sqrt{3}}}{1+\frac{r+1}{\sqrt{3}}\cdot \frac{r}{\sqrt{3}}}\right)$
$=\sum _{r=1}^{\infty }[{\tan }^{-1}\frac{r+1}{\sqrt{3}}-{\tan }^{-1}\frac{r}{\sqrt{3}}]$
$=\left(\underset{r\to \infty }{lim}{\tan }^{-1}\frac{r+1}{\sqrt{3}}\right)-{\tan }^{-1}\frac{1}{\sqrt{3}}$
$=\frac{\pi }{2}-\frac{\pi }{6}$
$=\frac{\pi }{3}$