Question

The A.M (Arithmetic Mean) of ${}^n{C_0},{}^n{C_1},{}^n{C_2},..........,{}^n{C_n}$ is
  \[
  A.{\text{ }}\dfrac{{{2^n}}}{n} \\
  B.{\text{ }}\dfrac{{{2^{n + 1}}}}{n} \\
  C.{\text{ }}\dfrac{{{2^n}}}{{n + 1}} \\
  D.{\text{ }}\dfrac{{{2^{n + 1}}}}{{n + 1}} \\
 \]

Answer 1

Hint: In order to find the arithmetic mean of the given terms, we will first use the basic formula for finding the arithmetic mean which is given as the sum of all the terms over the number of terms, further in order to find the sum of the given terms we will use the binomial theorem of series expansion with some consideration.

Complete step-by-step answer:
Given terms are: ${}^n{C_0},{}^n{C_1},{}^n{C_2},..........,{}^n{C_n}$
To find the A.M = Arithmetic Mean of the given terms.
Since the counting of the terms starts from 0 and ends at n. so, the number of terms in this series is n+1.
As we know that arithmetic mean is given as:
${\text{A}}{\text{.M}}{\text{.}} = \dfrac{{{\text{Sum of elements}}}}{{{\text{Number of elements}}}}$
So, for the above case we have:
${\text{A}}{\text{.M}}{\text{.}} = \dfrac{{{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n}}}{{n + 1}} - - - - - - (1)$
Now, let us find the sum of the terms in the numerator.
As we know that the binomial expansion of the term ${\left( {1 + x} \right)^n}$ is given as,
\[{\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + .......... + {}^n{C_n}{x^n}\]
Let us substitute x =1 in the expansion, it becomes
  \[
  \because {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + .......... + {}^n{C_n}{x^n} \\
   \Rightarrow {\left( {1 + 1} \right)^n} = {}^n{C_0} + {}^n{C_1}1 + {}^n{C_2}{1^2} + .......... + {}^n{C_n}{1^n} \\
   \Rightarrow {2^n} = {}^n{C_0} + {}^n{C_1}1 + {}^n{C_2}1 + .......... + {}^n{C_n}1 \\
   \Rightarrow {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n} - - - - - - - - (2) \\
 \]
Let us substitute the value from equation (2) into equation (1)
$
  \because {\text{A}}{\text{.M}}{\text{.}} = \dfrac{{{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n}}}{{n + 1}} \\
   \Rightarrow {\text{A}}{\text{.M}}{\text{.}} = \dfrac{{{2^n}}}{{n + 1}} \\
 $
Hence the A.M of ${}^n{C_0},{}^n{C_1},{}^n{C_2},..........,{}^n{C_n}$ is $\dfrac{{{2^n}}}{{n + 1}}$ .
Option C is the correct answer.

Note: The key in solving such types of problems is to know the concepts and formula of AM and binomial expansions to simplify the required values. Finding out the total number of elements in the given series is ‘n+1’ is the key step in solving this problem. While counting we have to remember that there are n terms and there is the 0th term, which makes it n+1 terms.