The AM of \( \Large ^{2n+1}C_{0},\ ^{2n+1}C_{2},\ .....^{2n+1}C_{n} \) is
Correct Answer: |
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D) \( \Large \frac{2^{2n}}{ \left(n+1\right) } \) |
Description for Correct answer:
\( \Large ^{2n+1}C_{0}=^{2n+1}C_{2n+1}=^{2n+1}C_{1}=^{2n+1}C_{2n}...^{2n+1}C_{r}=^{2n+1}C_{2n-r+1}=2^{2n+1} \)
Now, \( \Large ^{2n+1}C_{0}=^{2n+1}C_{2n+1}=^{2n+1}C_{1}=^{2n+1}C_{2n}...^{2n+1}C_{r}=^{2n+1}C_{2n-r+1}=2^{2n+1} \)
So, sum of first \( \Large \left(n+1\right) \) terms
= sum of last \( \Large \left(n+1\right) \) terms
=> \( \Large ^{2n+1}C_{0}+^{2n+1}C_{1}+^{2n+1}C_{2}+...+^{2n+1}C_{n}=2^{2n} \)
=> \( \Large \frac{^{2n+1}C_{0}+^{2n+1}C_{1}+^{2n+1}C_{2}+...+^{2n+1}C_{n}}{n+1} = \frac{2^{2n}}{n+1} \)
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