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If ${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}}$ and ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{^{n}{{C}_{r}}}}$, then $\dfrac{{{t}_{n}}}{{{S}_{n}}}$ is equal to
(a) $\dfrac{n}{2}$
(b) $\dfrac{n}{2}-1$
(c) n – 1
(d) $\dfrac{2n-1}{2}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve this question, we will try to write one of the two given equations in the form of another. To do this, we have to perform some manipulations of equations. Moreover, we will also need the definition of $^{n}{{C}_{r}}$, which is defined as $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and n! = n(n – 1)(n – 2) … 1. We shall also keep in mind that $^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$. With all these things, we shall be able to find the value of $\dfrac{{{t}_{n}}}{{{S}_{n}}}$.

Complete step by step answer:
It is given to us that ${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}}$ and ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{^{n}{{C}_{r}}}}$.
First of all, we will consider ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{^{n}{{C}_{r}}}}$.
If we add and subtract n in the numerator, the value of the equation will not change. Thus, we will change the numerator r with n – (n – r).
The changed value of ${{t}_{n}}$ will be as follows:
$\begin{align}
  & \Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n-\left( n-r \right)}{^{n}{{C}_{r}}}} \\
 & \Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{^{n}{{C}_{r}}}-\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}} \\
\end{align}$
We know that $\sum{\left( a+b \right)=\sum{a}+\sum{b}}$.
We will apply this rule in the equation of ${{t}_{n}}$.
$\Rightarrow {{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{^{n}{{C}_{r}}}}-\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}}$
We know that the value of n is constant and thus can be taken out of the summation.
$\Rightarrow {{t}_{n}}=n\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}}-\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}}$
But it is given to us that ${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}}$.
$\Rightarrow {{t}_{n}}=n{{S}_{n}}-\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}}......\left( 1 \right)$
Now, let us expand the right hand side of ${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{r}{^{n}{{C}_{r}}}}$.
\[\Rightarrow {{t}_{n}}=\dfrac{0}{^{n}{{C}_{0}}}+\dfrac{1}{^{n}{{C}_{1}}}+\dfrac{2}{^{n}{{C}_{2}}}+...+\dfrac{n}{^{n}{{C}_{n}}}......\left( 2 \right)\]
Also, let us assume that ${{A}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}}$.
From the properties of $^{n}{{C}_{r}}$, we know that $^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$.
Thus, we will replace the $^{n}{{C}_{r}}$ in the denominator with ${}^{n}{{C}_{n-r}}$.
 $\Rightarrow {{A}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{n-r}}}}$
We will now expand the right hand side of ${{A}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{n-r}}}}$.
 $\begin{align}
  & \Rightarrow {{A}_{n}}=\dfrac{n}{^{n}{{C}_{n}}}+\dfrac{n-1}{^{n}{{C}_{n-1}}}+...+\dfrac{n-\left( n-1 \right)}{^{n}{{C}_{n-\left( n-1 \right)}}}+\dfrac{n-n}{^{n}{{C}_{n-n}}} \\
 & \Rightarrow {{A}_{n}}=\dfrac{n}{^{n}{{C}_{n}}}+\dfrac{n-1}{^{n}{{C}_{n-1}}}+...+\dfrac{1}{^{n}{{C}_{1}}}+\dfrac{0}{^{n}{{C}_{0}}}......\left( 3 \right) \\
\end{align}$
When we compare (2) and (3), we can observe first term of (2) is same as the last term of (3), second term of (2) is same as second last term of (3) and last term of (2) is same as the first term of (2). This means that (2) = (3)
$\Rightarrow {{t}_{n}}={{A}_{n}}$
Thus, we will replace $\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{^{n}{{C}_{r}}}}={{A}_{n}}$ with ${{t}_{n}}$ in the equation (1).
$\begin{align}
  & \Rightarrow {{t}_{n}}=n{{S}_{n}}-{{t}_{n}} \\
 & \Rightarrow 2{{t}_{n}}=n{{S}_{n}} \\
 & \Rightarrow \dfrac{{{t}_{n}}}{{{S}_{n}}}=\dfrac{n}{2} \\
\end{align}$

So, the correct answer is “Option A”.

Note: This question involves basic understanding of the summation and problem-solving method. Another simpler method to solve this question is take some arbitrary value for n and solve the problem numerically. The value of n must be greater than 0 and it must be so that every option is unique. For example, n cannot be 2, as with n = 2, option (a) and option (c) will be the same. The value of n can be 4, as with n = 4, every option yields different values.