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Q.
In the expansion of $ {{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{12}} $ , the term independent of $ x $ is
Jharkhand CECEJharkhand CECE 2008
Solution:
The general term in the expansion of
$ {{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{12}} $ is
$ {{T}_{r+1}}{{=}^{12}}{{C}_{r}}{{(2{{x}^{2}})}^{12-r}}\cdot {{\left( -\frac{1}{x} \right)}^{r}} $
$ ={{(-1)}^{r}}^{12}{{C}_{r}}{{2}^{12-r}}\cdot {{x}^{24-2r-r}} $
$ ={{(-1)}^{r}}^{12}{{C}_{r}}\cdot {{2}^{12-r}}\cdot {{x}^{24-3r}} $
The term independent of $ x $ , if $ 24-3r=0 $
$ \Rightarrow $ $ 24=3r $
$ \Rightarrow $ $ r=8 $
Now, $ r+1=8+1 $ $ =9th $ term