First slide

Introduction to statistics

Question

The AM of $^{2 n + 1} C_{0},^{2 n + 1} C_{1},^{2 n + 1} C_{2}, \dots,^{2 n + 1} C_{n}$ is

Difficult

A

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

$\frac{2^{n}}{n + 1}$
C

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

$\frac{2^{2 n}}{(n + 1)}$

Solution

$\begin{aligned} ^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + \dots +^{2 n + 1} C_{2 n} +^{2 n + 1} C_{2 n + 1} \\ = 2^{2 n + 1} \end{aligned}$
Now $\begin{aligned} ^{2 n + 1} C_{0} =^{2 n + 1} C_{2 n + 1}, \\ ^{2 n + 1} C_{1} =^{2 n + 1} C_{2 n} \dots^{2 n + 1} C_{r} =^{2 n + 1} C_{2 n - r + 1} \end{aligned}$
So, sum of first (n + 1 ) terms
= sum of last (n + 1 ) terms
$\begin{array}{l} \Rightarrow^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + \dots +^{2 n + 1} C_{n} = 2^{2 n} \\ \Rightarrow \frac{^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + \dots +^{2 n + 1} C_{n}}{n + 1} = \frac{2^{2 n}}{(n + 1)} \end{array}$

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