Question

If $\tan \left(\theta \right)=\sqrt{\frac{3}{2}}$ the sum of the infinite series $1+2(1-\cos (\theta \left)\right)+3(1-\cos {\left(\theta \right))}^2+4(1-\cos {\left(\theta \right))}^3+..............\infty$ is

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

Answer 1

The correct option is D

$\frac{5}{2}$

Explanation for the correct option:

Given: $\tan \left(\theta \right)=\sqrt{\frac{3}{2}}$

Let $S=1+2(1-\cos (\theta \left)\right)+3(1-\cos {\left(\theta \right))}^2+4(1-\cos {\left(\theta \right))}^3+..............\infty$ ------(1)

multiplying both sides by $1-\cos \left(\theta \right)$

$$\begin{gathered} \Rightarrow S(1-\cos (\theta \left)\right)=(1-\cos (\theta \left)\right)+2(1-\cos {\left(\theta \right))}^2+3(1-\cos {\left(\theta \right))}^3+4(1-\cos {\left(\theta \right))}^4+..............\infty \\ \Rightarrow S-S\cos (\theta )=(1-\cos \left(\theta \right))+2(1-\cos {\left(\theta \right))}^2+3(1-\cos {\left(\theta \right))}^3+4(1-\cos {\left(\theta \right))}^4+..............\infty \end{gathered}$$

multiplying both sides by $-1$

$\Rightarrow S\cos \left(\theta \right)-S=-(1-\cos (\theta \left)\right)-2(1-\cos {\left(\theta \right))}^2-3(1-\cos {\left(\theta \right))}^3-4(1-\cos {\left(\theta \right))}^4+..............\infty$

add equation (1)

$$\begin{gathered} \Rightarrow S\cos \left(\theta \right)-S+S=\left[-(1-\cos (\theta \left)\right)-2(1-\cos {\left(\theta \right))}^2-3(1-\cos {\left(\theta \right))}^3-4(1-\cos {\left(\theta \right))}^4+..............\infty \right]+\left[1+2(1-\cos (\theta \left)\right)+3(1-\cos {\left(\theta \right))}^2+.............\infty \right] \\ \Rightarrow S\cos \left(\theta \right)-S+S=\left[-(1-\cos (\theta \left)\right)-2(1-\cos {\left(\theta \right))}^2-3(1-\cos {\left(\theta \right))}^3-4(1-\cos {\left(\theta \right))}^4+..............\infty \right]+\left[1+2(1-\cos (\theta \left)\right)+3(1-\cos {\left(\theta \right))}^2+.............\infty \right] \\ \Rightarrow S\cos \left(\theta \right)=1+(1-\cos (\theta \left)\right)+(1-\cos {\left(\theta \right))}^2+.............\infty \end{gathered}$$This equation is a geometric progression with the first term $a=1$and common ratio $r=1-\cos \left(\theta \right)$

sum of an infinite geometric progression $=\frac{a}{1-r}$

$$\begin{gathered} \Rightarrow S\cos \left(\theta \right)=\frac{1}{1-\left(1-\cos \left(\theta \right)\right)} \\ \Rightarrow S\cos \left(\theta \right)=\frac{1}{1-1+\cos \left(\theta \right)} \\ \Rightarrow S\cos \left(\theta \right)=\frac{1}{\cos \left(\theta \right)} \\ \Rightarrow S=\frac{1}{{\cos }^2\left(\theta \right)} \\ \Rightarrow S={\sec }^2\left(\theta \right)\left[\frac{1}{cos\left(x\right)}=sec\left(x\right)\right] \\ \Rightarrow S={\tan }^2\left(\theta \right)+1\left[se{c}^2\left(x\right)=ta{n}^2\left(x\right)+1\right] \\ \Rightarrow S={\left(\sqrt{\frac{3}{2}}\right)}^2+1 \\ \Rightarrow S=\frac{3}{2}+1 \\ \Rightarrow S=\frac{5}{2} \end{gathered}$$

Hence, option D is correct.