A) \[\frac{1\cdot 3\cdot 5...(2n-1)}{n!}\]
B) \[\frac{1\cdot 3\cdot 5...(2n-1)}{n!}{{2}^{n}}\]
C) \[\frac{1\cdot 3\cdot 5...(2n+1)}{(n+1)!}\]
D) None of the above
Correct Answer: C
Solution :
Let \[(r+1)th\] term is independent of \[x\] in the expansion of\[{{\left( x+\frac{1}{x} \right)}^{2n}}\]. \[{{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\times }^{2n-2r}}\] This term is independent of\[x\]. \[\therefore \] \[2n-2r=0\] \[\Rightarrow \] \[r=n\] Thus, \[{{T}_{r+1}}{{=}^{2n}}{{C}_{n}}\] \[=\frac{(2n)!}{n!n!}=\frac{1\cdot 2\cdot 3\cdot 4...(2n-1)(2n)}{n!n!}\] \[=\frac{(1\cdot 3\cdot 5...(2n-1)(1\cdot 2...n){{2}^{n}}}{n!n!}\] \[=\frac{1\cdot 3\cdot 5...(2n-1)}{n!}\cdot {{2}^{n}}\]
You need to login to perform this action.
You will be redirected in
3 sec
