Question

If $\tan \theta = \sqrt {\dfrac{3}{2}} $ ,then the sum of the infinite series $1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty $ is
A) $\dfrac{2}{3}$
B) $\dfrac{{\sqrt 3 }}{4}$
C) $\dfrac{5}{{2\sqrt 2 }}$
D) $\dfrac{5}{2}$

Answer 1

Hint: In the above question we have been given the value of $\tan \theta $ . We will first assume the given infinite series as
$S$ i.e. $S = 1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty $.
After that we will multiply the expression with $(1 - \cos \theta )$ .
The general form of the infinite geometric series is $a,a{r^2},a{r^3}...a{r^n}$ .
By comparing the given series in the question, we can see that the value of $r = (1 - \cos \theta )$ .
We will apply the formula of sum of infinites series i.e.
${S_\infty } = \dfrac{a}{{1 - r}}$ , where $a$ is the first term i.e. $1$
and $r$ is the common ratio, $(1 - \cos \theta )$ .

Complete step by step answer:
We have been given the infinite geometric series $1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty $ .
Let us assume that $S = 1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty $ .
This is our first equation.
We will multiply both the side of the equation with $(1 - \cos \theta )$
$S(1 - \cos \theta ) = (1 - \cos \theta )\left\{ {1 + 2(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2} + 4{{(1 - \cos \theta )}^3} + ...\infty } \right\}$ .
We can write the above also as $S - S\cos \theta = \left\{ {1(1 - \cos \theta ) + 2(1 - \cos \theta )(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2}(1 - \cos \theta ) + 4{{(1 - \cos \theta )}^3}(1 - \cos \theta ) + ...\infty } \right\}$On multiplying the above expression can be written as $S - S\cos \theta = \left\{ {(1 - \cos \theta ) + 2{{(1 - \cos \theta )}^2} + 3{{(1 - \cos \theta )}^3} + 4{{(1 - \cos \theta )}^4} + ...\infty } \right\}$
This is our second equation.
Now we will subtract second equation from first equation i.e.
$S - \left( {S - S\cos \theta } \right) = \left\{ {1 + 2(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2} + 4{{(1 - \cos \theta )}^3} + ...\infty } \right\} - \left\{ {(1 - \cos \theta ) + 2{{(1 - \cos \theta )}^2} + 3{{(1 - \cos \theta )}^3} + 4{{(1 - \cos \theta )}^4} + ...\infty } \right\}$By breaking the values we can write: $S - S + S\cos \theta = 1 + 2(1 - \cos \theta ) - (1 - \cos \theta ) + 3{(1 - \cos \theta )^2} - 2{(1 - \cos \theta )^2} + ...\infty $
On subtracting the values above we have
 $S\cos \theta = 1 + (1 - \cos \theta ) + {(1 - \cos \theta )^2} + ...\infty $
Now here we have the infinite series. So we will apply the formula of sum of the infinite series i.e.
${S_\infty } = \dfrac{a}{{1 - r}}$ , where $a$ is the first term i.e.
 $1$ and,
 $r$ is the common ratio,
$(1 - \cos \theta )$
By putting the values in the formula we have
$S\cos \theta = \dfrac{1}{{1 - (1 - \cos \theta )}}$
We can break down the value and it gives us
$\dfrac{1}{{1 - 1 + \cos \theta }} = \dfrac{1}{{\cos \theta }}$
By can transfer the cosine function to the other side i.e.
$S = \dfrac{1}{{\cos \theta \times \cos \theta }} = \dfrac{1}{{{{\cos }^2}\theta }}$
We know that
$\dfrac{1}{{\cos \theta }} = \sec \theta $
So by putting this we can write
 $S = {\sec ^2}\theta $
WE know the trigonometric identity that
${\sec ^2}\theta = 1 + {\tan ^2}\theta $, we will put this in the expression
We have,
$S = 1 + {\tan ^2}\theta $,
And the value of $\tan \theta $ is given, so we will put that in the expression:
$S = 1 + {\left( {\sqrt {\dfrac{3}{2}} } \right)^2}$
On simplifying we have
$1 + \dfrac{3}{2} = \dfrac{{2 + 3}}{2}$
It gives us the value
$S = \dfrac{5}{2}$
Therefore, If $\tan \theta = \sqrt {\dfrac{3}{2}} $ ,then the sum of the infinite series $1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty $ is $\dfrac{5}{2}$. Hence the correct option is (D).

Note:
We should know that in the above we have been given to find sum of infinite series, i.e.
here $n$ goes to $\infty $
 If we have to find the sum of finite geometric series is
${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$ , where $n$ is the number of terms.