A) \[4\]
B) \[6\]
C) \[7\]
D) \[9\]
Correct Answer: B
Solution :
[b] We have \[{{\left( {{x}^{2}}-\frac{1}{x} \right)}^{n}}\] \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{({{x}^{2}})}^{n-r}}{{(-1)}^{r}}{{x}^{-r}}{{=}^{n}}{{C}_{r}}\,{{x}^{2n-3r}}{{(-1)}^{r}}\] Constant term \[{{=}^{n}}{{C}_{r}}{{(-1)}^{r}}\] if \[2n=3r\] \[\Rightarrow \,\,{{\,}^{n}}{{C}_{2n/3}}{{(-1)}^{2n/3}}=15{{=}^{6}}{{C}_{4}}\] \[\Rightarrow \,\,\,n=6\]You need to login to perform this action.
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