A) 0
B) \[{{2.}^{9}}{{C}_{4}}\]
C) \[^{9}{{C}_{5}}\]
D) \[^{9}{{C}_{3}}\]
Correct Answer: D
Solution :
\[{{T}_{r+1}}\,\] in \[{{\left( {{x}^{2}}-\frac{1}{x} \right)}^{9}}\] is \[^{9}{{C}_{r}}{{x}^{2(9-r)}}\,{{\left( \frac{-1}{x} \right)}^{r}}\] \[{{=}^{9}}{{C}_{r}}\,.{{x}^{18-3r}}\,.{{(-1)}^{r}}\] For term impendent of \[x,\,\,18-3r=0\Rightarrow \,r=6\] \[\therefore \] 7th term is independent of x and equals \[^{9}{{C}_{6}}\,{{=}^{9}}{{C}_{3}}\,=84\] Also there are 10 terms, hence 5th term and 6th and the two middle term \[{{T}_{5}}\,{{=}^{9}}{{C}_{4}}.{{x}^{6}}\] \[{{T}_{6}}\,={{-}^{9}}{{C}_{5}}.{{x}^{3}}\] \[\therefore \,\,q=\] Coefficient of 5th + coefficient of 6th term \[{{=}^{9}}{{C}_{4}}\,{{-}^{9}}{{C}_{5}}=0\] Hence, p =84; q = 0 \[\therefore \,\,p-q\,{{=}^{9}}{{C}_{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
