A) \[\frac{(2n+1)!}{n!(n+1)!}\]
B) \[\frac{(2n+2)!}{n!(n+1)!}\]
C) \[\frac{(2n+1)!}{{{[(n+1)!]}^{2}}}\]
D) \[\frac{(2n)!}{{{(n!)}^{2}}}\]
Correct Answer: A
Solution :
[a]: We know, \[\frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{N-r+1}{r}.x\] Given\[N=2n+1\Rightarrow \frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{2n+2-r}{r}.x\] \[\therefore \]\[{{T}_{r+1}}\ge {{T}_{r}}\] \[\Rightarrow \]\[2n+2-r\ge r\Rightarrow 2n+2\ge 2r\Rightarrow r\le n+1\] \[\therefore \]\[r=n\] \[{{T}_{r+1}}={{T}_{n+1}}{{=}^{2n+1}}{{C}_{n+1}}=\frac{(2n+1)!}{(n+1)!n!}.\]
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