Question

The arithmetic mean of $ ^{n}{{C}_{0}} $ , $ ^{n}{{C}_{1}} $ , $ ^{n}{{C}_{2}} $ …., $ ^{n}{{C}_{n}} $ is
(a) $ \dfrac{{{2}^{n}}}{n+1} $
(b) $ \dfrac{{{2}^{n}}}{n} $
(c) $ \dfrac{{{2}^{n-1}}}{n+1} $
(d) $ \dfrac{{{2}^{n-1}}}{n} $
(e) $ \dfrac{{{2}^{n+1}}}{n} $

Answer 1

Hint: To begin with, we will define the arithmetic mean and give an example to understand it better. Then, to solve this question, first of all we will find the value of $ ^{n}{{C}_{0}} $ + $ ^{n}{{C}_{1}} $ + $ ^{n}{{C}_{2}} $ …. + $ ^{n}{{C}_{n}} $ with the help of concepts from binomial expansion. Once we get the value of $ ^{n}{{C}_{0}} $ + $ ^{n}{{C}_{1}} $ + $ ^{n}{{C}_{2}} $ …. + $ ^{n}{{C}_{n}} $ , we can divide it by the number of terms to get the arithmetic mean of $ ^{n}{{C}_{0}} $ , $ ^{n}{{C}_{1}} $ , $ ^{n}{{C}_{2}} $ …., $ ^{n}{{C}_{n}} $ .

Complete step-by-step answer:
In mathematics, arithmetic mean is defined as the quotient of the sum of the values of all the terms present in the sample set and the number of elements in the sample set.
Suppose X is a sample set defined as X : {2, 6, 7, 2, 1, 6, 9}.
The number of elements in the sample set is n(X) = 7
Now, the sum of the values of the elements in the set is (2 + 6 + 7 + 2 + 1 + 6 + 9) = 33
Thus, the arithmetic mean will be $ \overline{X}=\dfrac{33}{7} $ = 4.714
Now, coming to our problem, we know that the binomial expansion of $ {{\left( 1+x \right)}^{n}} $ is given as follows:
 $ {{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+....{{+}^{n}}{{C}_{n}}{{x}^{n}} $ .
Now, we shall substitute x = 1 in this expansion.
 $ \begin{align}
  & \Rightarrow {{\left( 1+1 \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}\left( 1 \right){{+}^{n}}{{C}_{2}}{{\left( 1 \right)}^{2}}{{+}^{n}}{{C}_{3}}{{\left( 1 \right)}^{3}}+....{{+}^{n}}{{C}_{n}}{{\left( 1 \right)}^{n}} \\
 & \Rightarrow {{\left( 2 \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{3}}+....{{+}^{n}}{{C}_{n}} \\
\end{align} $
Thus, the value of $ ^{n}{{C}_{0}} $ + $ ^{n}{{C}_{1}} $ + $ ^{n}{{C}_{2}} $ …. + $ ^{n}{{C}_{n}} $ = $ {{2}^{n}} $
As we can see, the series has (n + 1) elements.
Therefore, the arithmetic means of the series will be the sum of all the terms divided by the number of terms.
 $ \Rightarrow $ AM = $ \dfrac{^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{3}}+....{{+}^{n}}{{C}_{n}}}{n+1} $
 $ \Rightarrow $ AM = $ \dfrac{{{2}^{n}}}{n+1} $
Therefore, the arithmetic mean is $ \dfrac{{{2}^{n}}}{n+1} $ .
So, the correct answer is “Option A”.

Note: The prerequisites for this question is the student has to know about binomial expansions. Also, they don’t have to prove that the value $ ^{n}{{C}_{0}} $ + $ ^{n}{{C}_{1}} $ + $ ^{n}{{C}_{2}} $ …. + $ ^{n}{{C}_{n}} $ = $ {{2}^{n}} $ every time. This can be remembered as a standard solution.