• # question_answer The coefficient of ${{x}^{10}}$ in the expansion of${{\left( 3{{x}^{2}}-\frac{1}{{{x}^{2}}} \right)}^{15}}$is A) $\frac{15!}{10!5!}{{3}^{10}}$                   B) $-\frac{15!}{10!5!}{{3}^{10}}$C) $-\frac{15!{{3}^{5}}}{10!5!}$                    D) $-\frac{15!}{7!8!}{{3}^{8}}$

${{\left( 3{{x}^{2}}-\frac{1}{{{x}^{2}}} \right)}^{15}}$ ${{T}_{r+1}}{{=}^{15}}{{C}_{r}}{{(3{{x}^{2}})}^{15-r}}{{\left( -\frac{1}{{{x}^{2}}} \right)}^{r}}$        ${{=}^{15}}{{C}_{r}}{{3}^{15-r}}{{(-1)}^{r}}{{x}^{30-2r-2r}}$ Therefore$,$$30-4r=10\Rightarrow r=5$. Therefore,${{T}_{6}}={{-}^{15}}{{C}_{5}}{{3}^{10}}=\frac{-15!}{10!5!}{{3}^{10}}$