The AM of  2n+1C0,2n+1C1,2n+1C2,…,2n+1Cn is

# The AM of  is

1. A

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

2. B

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

3. C

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

4. D

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

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### Solution:

Now

So, sum of first (n + 1 ) terms

= sum of last (n + 1 ) terms

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