APPLYING \[{{T}_{r+1}}={{\,}^{N}}{{C}_{r}}{{X}^{n-r}}{{a}^{r}}\]for \[{{(x+a)}^{n}}\] Hence \[{{T}_{6}}={{\,}^{10}}{{C}_{5}}{{(2{{x}^{2}})}^{5}}{{\left( -\frac{1}{3{{x}^{2}}} \right)}^{5}}\] \[=-\frac{10\,!}{5\,!\,5\,!}32\times \frac{1}{243}=-\frac{896}{27}\] "> APPLYING \[{{T}_{r+1}}={{\,}^{N}}{{C}_{r}}{{X}^{n-r}}{{a}^{r}}\]for \[{{(x+a)}^{n}}\] Hence \[{{T}_{6}}={{\,}^{10}}{{C}_{5}}{{(2{{x}^{2}})}^{5}}{{\left( -\frac{1}{3{{x}^{2}}} \right)}^{5}}\] \[=-\frac{10\,!}{5\,!\,5\,!}32\times \frac{1}{243}=-\frac{896}{27}\] ">
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\[{{6}^{th}}\]term in expansion of \[{{\left( 2{{x}^{2}}-\frac{1}{3{{x}^{2}}} \right)}^{10}}\] is 

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 11 months ago

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APPLYING \[{{T}_{r+1}}={{\,}^{N}}{{C}_{r}}{{X}^{n-r}}{{a}^{r}}\]for \[{{(x+a)}^{n}}\] Hence \[{{T}_{6}}={{\,}^{10}}{{C}_{5}}{{(2{{x}^{2}})}^{5}}{{\left( -\frac{1}{3{{x}^{2}}} \right)}^{5}}\] \[=-\frac{10\,!}{5\,!\,5\,!}32\times \frac{1}{243}=-\frac{896}{27}\]
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