Question

The mean of the values 0, 1, 2,...... n having corresponding weights, \[{}^n{C_0},{}^n{C_1},.......{}^n{C_n}\] respectively is
$\left( a \right)\dfrac{{n + 1}}{2}$
$\left( b \right)\dfrac{{{2^{n - 1}} + 1}}{{n + 1}}$
$\left( c \right)\dfrac{{{2^n}}}{n}$
$\left( d \right)\dfrac{n}{2}$

Answer 1

Hint: In this particular question use the concept that the mean of the values is the ratio of the product of sum of values and their corresponding weights to the sum of all the weights, and use the property of Binomial expansion so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given values
0, 1, 2, ................., n
Corresponding weights \[{}^n{C_0},{}^n{C_1},.......{}^n{C_n}\].
Now we have to find out the mean of the values.
So the mean of the values is the ratio of the product of sum of values and their corresponding weights to the sum of all the weights.
Let the mean of the values be denoted by $\bar x$.
$ \Rightarrow \bar x = \dfrac{{\left( {0.{}^n{C_0}} \right) + \left( {1.{}^n{C_1}} \right) + \left( {2.{}^n{C_2}} \right) + ....... + \left( {n.{}^n{C_n}} \right)}}{{{}^n{C_0} + {}^n{C_1} + ...... + {}^n{C_n}}}$
Now the above equation is also written as,
$ \Rightarrow \bar x = \dfrac{{\sum\limits_{r = 0}^n {r.{}^n{C_r}} }}{{\sum\limits_{r = 0}^n {{}^n{C_r}} }}$
Now as we know that $\sum\limits_{r = 0}^n {r.{}^n{C_r}} = \sum\limits_{r = 1}^n {r.\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} $, $\left[ {\because {}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} \right]$ so use this property in the above equation we have,
$ \Rightarrow \bar x = \dfrac{{\sum\limits_{r = 1}^n {r.\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} }}{{\sum\limits_{r = 0}^n {{}^n{C_r}} }}$
Now simplify the above equation we have,
$ \Rightarrow \bar x = \dfrac{{\sum\limits_{r = 1}^n {n{}^{n - 1}{C_{r - 1}}} }}{{\sum\limits_{r = 0}^n {{}^n{C_r}} }}$
As n is independent of r, so n can be written outside the summation so we have,
$ \Rightarrow \bar x = \dfrac{{n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} }}{{\sum\limits_{r = 0}^n {{}^n{C_r}} }}$......................... (1)
Now as we know that according to the Binomial expansion the expansion of ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ...... + {}^n{C_n}{x^n}$
So in place of x substitute 1 we have,
$ \Rightarrow {\left( {1 + 1} \right)^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n}$
\[ \Rightarrow {\left( 2 \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} \]
Similarly, $\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} = {\left( 2 \right)^{n - 1}}$
So use these values in equation (1) we have,
$ \Rightarrow \bar x = \dfrac{{n{{\left( 2 \right)}^{n - 1}}}}{{{2^n}}}$
$ \Rightarrow \bar x = \dfrac{n}{{{2^{n - n + 1}}}} = \dfrac{n}{2}$
So this is the required mean of the given values.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that, $\sum\limits_{r = 0}^n {r.{}^n{C_r}} = \sum\limits_{r = 1}^n {r.\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} $, $\left[ {\because {}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} \right]$ and always recall the binomial expansion which is given as, ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ...... + {}^n{C_n}{x^n}$, therefore, ${\left( {1 + 1} \right)^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n}$.