A) \[\frac{15!}{10!5!}{{3}^{10}}\]
B) \[-\frac{15!}{10!5!}{{3}^{10}}\]
C) \[-\frac{15!{{3}^{5}}}{10!5!}\]
D) \[-\frac{15!}{7!8!}{{3}^{8}}\]
Correct Answer: B
Solution :
\[{{\left( 3{{x}^{2}}-\frac{1}{{{x}^{2}}} \right)}^{15}}\] \[{{T}_{r+1}}{{=}^{15}}{{C}_{r}}{{(3{{x}^{2}})}^{15-r}}{{\left( -\frac{1}{{{x}^{2}}} \right)}^{r}}\] \[{{=}^{15}}{{C}_{r}}{{3}^{15-r}}{{(-1)}^{r}}{{x}^{30-2r-2r}}\] Therefore\[,\]\[30-4r=10\Rightarrow r=5\]. Therefore,\[{{T}_{6}}={{-}^{15}}{{C}_{5}}{{3}^{10}}=\frac{-15!}{10!5!}{{3}^{10}}\]
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