Question

If $\sum\limits_{r = 0}^n {\dfrac{1}{{{}^n{C_r}}} = {S_n},} {\text{ }}{{\text{t}}_n} = \sum\limits_{r = 0}^n {\dfrac{r}{{{}^n{C_r}}}} ,$ then $\dfrac{{{t_n}}}{{{S_n}}} =$
A. $\dfrac{1}{4}n$
B. $\dfrac{1}{3}n$
C. $\dfrac{1}{2}n$
D. $n$

Answer 1

Hint : Here we will use the Combinations property and expansion of the equation by placing the limit. Find the correlation between the given terms and the required solution. Do simplification by using mathematical operations.

Complete step-by-step answer :
Given function- ${{\text{t}}_n} = \sum\limits_{r = 0}^n {\dfrac{r}{{{}^n{C_r}}}}$
The above function can be re-written as –substitute $r = n - (n - r)$ in the numerator part.
${{\text{t}}_n} = \sum\limits_{r = 0}^n {\dfrac{{n - (n - r)}}{{{}^n{C_r}}}}$
Split the denominator in the above equation –
${{\text{t}}_n} = \sum\limits_{r = 0}^n {\dfrac{n}{{{}^n{C_r}}}} - \sum\limits_{r = 0}^n {\dfrac{{(n - r)}}{{{}^n{C_r}}}}$
Now, take “n” common from the first term of the right hand side of the equation –
${{\text{t}}_n} = n\sum\limits_{r = 0}^n {\dfrac{1}{{{}^n{C_r}}}} - \sum\limits_{r = 0}^n {\dfrac{{(n - r)}}{{{}^n{C_r}}}}$
Now we are also given that $\sum\limits_{r = 0}^n {\dfrac{1}{{{}^n{C_r}}} = {S_n}}$ , substitute its value in the above equation –
${{\text{t}}_n} = n{S_n} - \sum\limits_{r = 0}^n {\dfrac{{(n - r)}}{{{}^n{C_r}}}}$
${{\text{t}}_n} = n{S_n} - {A_n}$ .... (A)
Let us consider
${A_n} = \sum\limits_{r = 0}^n {\dfrac{{(n - r)}}{{{}^n{C_r}}}}$
Here we should know the property that- ${}^n{C_r} = {}^n{C_{n - r}}$ , replace it in the above equation –
${A_n} = \sum\limits_{r = 0}^n {\dfrac{{(n - r)}}{{{}^n{C_{n - r}}}}}$
Expand the above equation and place the value of the summation which starts with $r = 0,1,....n$
${A_n} = \dfrac{n}{{{}^n{C_n}}} + \dfrac{{n - 1}}{{{}^n{C_{n - 1}}}} + \dfrac{{n - 2}}{{{}^n{C_{n - 2}}}} + .... + \dfrac{1}{{{}^n{C_1}}} + \dfrac{0}{{{}^n{C_0}}}$ .... (B)
Now take the given function –
${{\text{t}}_n} = \sum\limits_{r = 0}^n {\dfrac{r}{{{}^n{C_r}}}}$
Expand the above equation and place the value of the summation which starts with $r = 0,1,....n$
${t_n} = \dfrac{0}{{{}^n{C_0}}} + \dfrac{1}{{{}^n{C_1}}} + ........\dfrac{{n - 2}}{{{}^n{C_{n - 2}}}} + \dfrac{{n - 1}}{{{}^n{C_{n - 1}}}} + \dfrac{n}{{{}^n{C_n}}}$ .... (C)
Now, compare the equations (B) and (C), we observe that the summation of addition is in reverse order where the terms are the same.
Therefore by using additive property, $a + b = b + a$
We can say that –
${A_n} = {t_n}$
Place the above value in the equation (A)
${{\text{t}}_n} = n{S_n} - {t_n}$
When the term is moved from one side to another sign also changes. Positive sign changes to negative and vice-versa.
${{\text{t}}_n} + {{\text{t}}_n} = n{S_n}$
Simplify the above equation –
${\text{2}}{{\text{t}}_n} = n{S_n}$
When the term in multiplicative is moved from one side to another, it goes to the division on the opposite side and vice-versa. Arrange the above equation in the form of the required solution-
$\dfrac{{{{\text{t}}_n}}}{{{S_n}}} = \dfrac{n}{2}$
So, the correct answer is “Option C”.

Note : To split the given equation and rewrite as per the required solution is the trickiest part. Do wisely and carefully solve using basic expansion of the equations and compare the two equations to find the correlation to get the unknown ratio.