A) \[^{-15}{{C}_{3}}\]
B) \[^{15}{{C}_{4}}\]
C) \[^{-15}{{C}_{5}}\]
D) \[^{15}{{C}_{2}}\]
Correct Answer: B
Solution :
In \[{{\left( {{x}^{4}}-\frac{1}{{{x}^{3}}} \right)}^{15}}\] \[{{T}_{r+1}}={{\,}^{15}}{{C}_{r}}{{({{x}^{4}})}^{15-r}}{{\left( -\frac{1}{{{x}^{3}}} \right)}^{r}}\] \[={{\,}^{15}}{{C}_{r}}{{x}^{60-4r}}\frac{{{(-1)}^{r}}}{{{x}^{3r}}}\] \[={{\,}^{15}}{{C}_{r}}{{x}^{60-7r}}{{(-1)}^{r}}\] Now equating the power of \[x\] to 32, we get \[60-7r=32\] \[\therefore \] \[7r=60-32\] \[7r=28\] \[r=4\] Now coefficient of \[{{x}^{32}}\] in \[{{\left( {{x}^{4}}-\frac{1}{{{x}^{3}}} \right)}^{15}}\] \[={{\,}^{15}}{{C}_{4}}\,{{(-1)}^{4}}{{=}^{15}}{{C}_{4}}\]
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