Question

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) $\dfrac{1}{2}n$
(b) $\dfrac{1}{2}n-1$
(c) $n-1$
(d) $\dfrac{2n-1}{2}$

Answer 1

Hint: First write the 2 variables and their values. Now take the ${{t}_{n}}$ term and manipulate it to into the ${{S}_{n}}$ term into the ${{t}_{n}}$ term. Now find the value of the extra term also in terms of ${{t}_{n}}$ term. By this you get an equation in these 2 variables. Now try to group the same variable on the same side of the equation. By this you get the equation with variables on both sides. Now bring ${{S}_{n}}$ term to denominator and coefficient of ${{t}_{n}}$ to denominator of other side. By this you get the required ratio. This fraction will be your result.

Complete step-by-step answer:
First term given into the question, can be written as
${{S}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}}.........(i)$
The second term given in the question can be written as
${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{^{n}{{C}_{r}}}}.........(ii)$
By expanding this term above, we can write the equation:
${{t}_{n}}=\dfrac{0}{^{n}{{C}_{0}}}+\dfrac{1}{^{n}{{C}_{1}}}+............+\dfrac{n}{^{n}{{C}_{n}}}$
By reversing the terms, i.e., writing the terms in reverse order, we get:
${{t}_{n}}=\dfrac{n}{^{n}{{C}_{n}}}+\dfrac{n-1}{^{n}{{C}_{n-1}}}+............+\dfrac{1}{^{n}{{C}_{1}}}+\dfrac{0}{^{n}{{C}_{0}}}...........(iii)$
By general algebra, we can write term r as follows:
$r=n-\left( n-r \right)$
So, by substituting this into equation (ii), we get it as:
${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n-\left( n-r \right)}{^{n}{{C}_{r}}}}$
By general algebra we know that a fraction:
$\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}$
By applying this we can write it as follows here:
${{t}_{n}}-\sum\limits_{r=0}^{n}{\left( \dfrac{n}{^{n}{{C}_{r}}}-\dfrac{n-r}{^{n}{{C}_{r}}} \right)}$
By normal simplification, we get the values of above term:
${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{^{n}{{C}_{r}}}-}\sum\limits_{r=0}^{n}{\dfrac{n-r}{^{n}{{C}_{r}}}...........(iv)}$
By taking the second term separately, we get it as:
$\sum\limits_{r=0}^{n}{\dfrac{n-r}{^{n}{{C}_{r}}}}$
By general knowledge of combinations, we know the relation:
$^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$
By substituting the above equation, we can write it as:
$\sum\limits_{r=0}^{n}{\dfrac{n-r}{^{n}{{C}_{n-r}}}}$
By expanding this by all terms, we get it as:
$\dfrac{n}{^{n}{{C}_{n}}}+\dfrac{n-1}{^{n}{{C}_{n-1}}}+.............+\dfrac{1}{^{n}{{C}_{1}}}+\dfrac{0}{^{n}{{C}_{0}}}$
Be equation (ii), we can say that this expression is nothing but:
$\sum\limits_{r=0}^{n}{\dfrac{n-r}{^{n}{{C}_{r}}}}={{t}_{n}}.......(v)$
By substituting equation (v) in the equation (iv), we get:
${{t}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{n}{^{n}{{C}_{r}}}-}{{t}_{n}}$
By simplifying the above equation, we can write it as:
${{t}_{n}}=n\sum\limits_{r=0}^{n}{\dfrac{1}{^{n}{{C}_{r}}}-}{{t}_{n}}$
By substituting the equation (i) back into this equation:
${{t}_{n}}=nSn-{{t}_{n}}$
By adding the term ${{t}_{n}}$ on both sides, we get it as:
$2{{t}_{n}}=nSn$
By dividing with term ${{S}_{n}}$ on both sides, we get it as:
$\dfrac{2{{t}_{n}}}{{{S}_{n}}}=n$
By dividing with 2 on both sides of equation, we get it as:
$\dfrac{{{t}_{n}}}{{{S}_{n}}}=\dfrac{n}{2}$
Therefore, the required answer is $\dfrac{n}{2}$ for the given question. Option (a) is correct.

Note: The main idea in this question is to find the value of the extra term which is also equal to ${{t}_{n}}$. Whenever you see a summation you must look at it in all directions in your vision. If you miss this idea and solve normally by combinations your answer may lead to 5-6 pages. So, always try to see summation in all directions. It is a great trick.