Question

The mean of the numbers \[\dfrac{{{}^{50}{C_0}}}{1},\dfrac{{{}^{50}{C_2}}}{3},\dfrac{{{}^{50}{C_4}}}{5},....\dfrac{{{}^{50}{C_{50}}}}{{51}}\] equals:
A. \[\dfrac{{{2^{50}}}}{{51}}\]
B. \[\dfrac{{{2^{49}}}}{{51}}\]
C. \[\dfrac{{{2^{49}}}}{{39 \times 17}}\]
D. None of these

Answer 1

Hint: The given terms are in the form of combinations. The combinations are the concept that says a word or any item that can be placed or arranged in different ways. So here it is asked to find the mean, the mean is defined as the sum of the given numbers divided by the number of terms.

Complete step-by-step answer:
The terms are given to find the mean are \[\dfrac{{{}^{50}{C_0}}}{1},\dfrac{{{}^{50}{C_2}}}{3},\dfrac{{{}^{50}{C_4}}}{5},....\dfrac{{{}^{50}{C_{50}}}}{{51}}\]
Here, the numerator is in the form of\[{}^n{C_0}\], it means n will be 50.
Taking,
\[{(1 + x)^{50}} = {}^{50}{C_0} + {}^{50}{C_2}{x^1} + .... + {}^{50}{C_{50}}{x^{50}} \ldots \ldots \ldots \left( {\rm{i}} \right)\]
As we know, the equation is,
\[{(1 + x)^{50}} = {}^{50}{C_0} - {}^{50}{C_2}{x^1} + .... + {}^{50}{C_{50}}{x^{50}} \ldots \ldots \ldots \left( {{\rm{ii}}} \right)\]
Adding both equation 1 and 2, then we will get,
\[{(1 + x)^{50}} = {}^{50}{C_0}x - \dfrac{{{}^{50}{C_3}{x^3}}}{3} + .... + \dfrac{{{}^{50}{C_{50}}{x^{51}}}}{{51}}\],
Now integrating the things with the help of limits 0 to 1, by solving it we will get
\[\begin{array}{c}
{(1 + x)^{50}} = \dfrac{1}{2}{\left( {\dfrac{{{{(1 + x)}^{50}}}}{{51}} - \dfrac{{{{(1 - x)}^{50}}}}{{51}}} \right)_0}^1\\
 = \dfrac{1}{2} \times \dfrac{{{2^{51}}}}{{51}}\\
 = \dfrac{{{2^{50}}}}{{51}}
\end{array}\]
The equation required to find the mean of the following terms is
We know that the mean is defined as the ratio of the sum of all numbers to the number of terms,
\[Mean{\rm{ }} = \dfrac{{{\rm{sum of the all numbers}}}}{{{\rm{number of terms}}}}\]
Then mean is
\[\begin{array}{l}
 = \dfrac{{{}^{50}{C_0} + {}^{50}{C_2}{x^1} + .... + {}^{50}{C_{50}}{x^{50}}}}{{26}}\\
 = \dfrac{{\dfrac{{{2^{50}}}}{{51}}}}{{26}}\\
 = \dfrac{{{2^{49}}}}{{39 \times 17}}
\end{array}\]
Therefore, the mean of the given numbers is\[\dfrac{{{2^{49}}}}{{39 \times 17}}\],
So, the correct answer is “Option C”.

Note: It should be noted that while the integration of the limits should be taken, they must be from 0 to 1. We should come to know the absolute meaning of the mean, its formula and how to work on it. Here we have to take the formula \[{(1 + x)^n}\]because the given terms can be framed according to it.