Q. The AM of $\ ^{2n + 1}C_{0},\ {\ ^{2n + 1}C}_{1},{\ ^{2n + 1}C}_{2},\ldots,{\ ^{2n + 1}C}_{n}$ is
A) $\frac{2^{n}}{n}$
B) $\frac{2^{n}}{n + 1}$
C) $\frac{2^{2n}}{n}$
D) $\frac{2^{2n}}{(n + 1)}$
Explanation
(d)
$\ ^{2n + 1}C_{0} + {\ ^{2n + 1}C}_{1} + {\ ^{2n + 1}C}_{2} + ... + {\ ^{2n + 1}C}_{2n} + {\ ^{2n + 1}C}_{2n + 1} = 2^{2n + 1}\$
Now, $\ ^{2n + 1}C_{0} = \ {\ ^{2n + 1}C}_{2n + 1},$
${\ ^{2n + 1}C}_{1} = {\ ^{2n + 1}C}_{2n}...{\ ^{2n + 1}C}_{r} = {\ ^{2n + 1}C}_{2n - r + 1}$
So, sum of first $(n + 1)$ terms= sum of last $(n + 1)$ terms
$\Rightarrow \ \ ^{2n + 1}C_{0} + {\ ^{2n + 1}C}_{1} + {\ ^{2n + 1}C}_{2} + ... + {\ ^{2n + 1}C}_{n} = 2^{2n}\$
$\Rightarrow \frac{\ ^{2n + 1}C_{0} + {\ ^{2n + 1}C}_{1} + {\ ^{2n + 1}C}_{2} + ... + {\ ^{2n + 1}C}_{n}}{n + 1} = \frac{2^{2n}}{(n + 1)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.