Question

If ${{\text{t}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $ and ${{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $, where ${\text{k}} \in {{\text{Z}}^ + }$, then ${\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right)$
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{2}$

Answer 1

- Hint: To solve this problem, we will use the property ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}}$. We will simplify \[{{\text{t}}_{\text{n}}}\] and \[{{\text{S}}_{\text{n}}}\] by using this property and also find a relation between them. Then we will use the trigonometric value of the inverse function ${\text{co}}{{\text{s}}^{ - 1}}{\text{x}}$ to find the value of ${\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right)$.

Complete step-by-step solution -

Now, we are given
${{\text{t}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $ and ${{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $
Now, r can be written as r = n – (n – r), Therefore, ${{\text{S}}_{\text{n}}}$ can be written as
${{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - (n - r)}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $
${{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n }}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} {\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $
${{\text{S}}_{\text{n}}}{\text{ = n}}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{1 }}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} {\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $
Therefore, ${{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}{\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $ … (1)
Now, let ${{\text{b}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $
From the property ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}}$, we get
${{\text{b}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - r}}}})}^{\text{k}}}}}} $
Expanding ${{\text{b}}_{\text{n}}}$, we get
${{\text{b}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{n}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{{\text{n - 1}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - 1}}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{{\text{n - 2}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - }}2}})}^{\text{k}}}}}{\text{ + }}........{\text{ + }}\dfrac{{\text{1}}}{{{{({}^{\text{n}}{{\text{C}}_1})}^{\text{k}}}}}{\text{ + }}\dfrac{0}{{{{({}^{\text{n}}{{\text{C}}_0})}^{\text{k}}}}}$
Also, ${{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} $.So, expanding ${{\text{S}}_{\text{n}}}$, we get
${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{0}{{{{({}^{\text{n}}{{\text{C}}_0})}^{\text{k}}}}}{\text{ + }}\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_1})}^{\text{k}}}}}{\text{ + }}\dfrac{2}{{{{({}^{\text{n}}{{\text{C}}_2})}^{\text{k}}}}}{\text{ + }}........{\text{ + }}\dfrac{{{\text{n - 1}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - 1}}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{\text{n}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{n}}})}^{\text{k}}}}}$
Now, we can see that both the expansion of ${{\text{b}}_{\text{n}}}$ and ${{\text{S}}_{\text{n}}}$ are equal. Therefore, we get
${{\text{S}}_{\text{n}}}$ = ${{\text{b}}_{\text{n}}}$. So, from equation (1), we get
${{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}{\text{ - }}{{\text{S}}_{\text{n}}}$
${\text{2}}{{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}$
Now, ${\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right)$ = ${\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{2}}{{\text{S}}_{\text{n}}}}}} \right)$ = ${\cos ^{ - 1}}\left( {\dfrac{1}{{\text{2}}}} \right)$. Now ${\cos ^{ - 1}}\left( {\dfrac{1}{{\text{2}}}} \right)$ = $\dfrac{\pi }{3}$
Therefore, ${\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right)$ = $\dfrac{\pi }{3}$
So, option (C) is correct.

Note: When we come up with such types of questions, we have to first simplify the given terms. Then, we have to find a relation between given terms as the question can be solved easily with the help of the relation. We have to use the property ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}}$ because it plays the most important role in finding the relation and solving the given problem.