A) 330
B) 42240
C) 114050
D) \[-32190\]
Correct Answer: B
Solution :
General term in the expansion of\[{{\left( 2{{x}^{4}}-\frac{1}{{{x}^{7}}} \right)}^{11}}\]is \[{{T}_{r+1}}{{=}^{11}}{{C}_{r}}{{(2{{x}^{4}})}^{11-r}}{{\left( -\frac{1}{{{x}^{7}}} \right)}^{r}}\] \[{{=}^{11}}{{C}_{r}}{{2}^{11-r}}{{(-1)}^{r}}{{x}^{4(11-r)-7r}}\] \[={{(-1)}^{r}}^{11}{{C}_{r}}{{2}^{11-r}}{{c}^{44-11r}}\] For independent of \[x,\text{ }44-11r=0\] \[\Rightarrow \] \[11r=44\] \[\Rightarrow \] \[r=4\] Thus, \[\therefore \] \[{{T}_{5}}={{(-1)}^{{{4}^{11}}}}{{C}_{4}}{{2}^{11-4}}\] \[=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}.{{(2)}^{7}}\] \[=330\times 128=42240\]
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