A) \[16\,{{\,}^{8}}{{C}_{4}}\]
B) \[^{8}{{C}_{4}}\]
C) \[^{8}{{C}_{5}}\]
D) none of these
Correct Answer: A
Solution :
\[\because \] Sum of coefficient of the expansion \[{{\left( \frac{1}{x}+2x \right)}^{n}}=6561\] \[\therefore \] \[{{(1+2)}^{n}}={{3}^{8}}\] \[\Rightarrow \] \[{{3}^{n}}={{3}^{8}}\Rightarrow n=8.\] Let \[(r+l)\]term is independent of\[x\]. \[\therefore \] \[{{T}_{r+1}}={{\,}^{8}}{{C}_{r}}{{\left( \frac{1}{x} \right)}^{r}}{{(2x)}^{8-r}}\] \[\Rightarrow \] \[{{\,}^{8}}{{C}_{r}}{{2}^{8-r}}{{x}^{8-2r}}\] Since this term is independent of \[x,\]then \[8-2r=0\] \[\Rightarrow \] \[r=4\] \[\therefore \]Coefficient of \[{{T}_{5}}={{\,}^{8}}{{C}_{4}}{{.2}^{4}}\] \[=16{{\,}^{8}}C{{\,}_{4}}.\]
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