Question

The median of ${}^{2n}C_0,{}^{2n}C_1,{}^{2n}C_2,{}^{2n}C_3,.....,{}^{2n}C_n$ (where $n$ is even)

• A. ${}^{2n}C_{\frac{n}{2}}$
• B. ${}^{2n}C_{\frac{n+1}{2}}$
• C. ${}^{2n}C_{\frac{n-1}{2}}$
• D. None of these

Answer 1

The correct option is A

${}^{2n}C_{\frac{n}{2}}$

Explanation for the correct option.

Find the median of the given series.

A series ${}^{2n}C_0,{}^{2n}C_1,{}^{2n}C_2,{}^{2n}C_3,.....,{}^{2n}C_n$ where $n$ is even.

Since, the number of terms in the series is $n+1$ which is odd.

Also, the mid-term is given by $\frac{n}{2}+1$.

So, the median $\text{M}$ of the given series is as follows:

$$\begin{gathered} \text{M}={}^{2n}C_{\frac{n}{2}+1-1} \\ \Rightarrow \text{M}={}^{2n}C_{\frac{n}{2}} \end{gathered}$$

Therefore, the median of the given series is ${}^{2n}C_{\frac{n}{2}}$.

Hence, option $A$ is the correct answer.