Question 11

Maths Question: 

$\text{Find the term independent of }x\text{ in the expansion }{{\left( 2x-\frac{3}{{{x}^{2}}} \right)}^{6}}$

Maths Solution: 

$\begin{align}  & \text{The general term of expansion will be }{{\left( 2x-\frac{3}{{{x}^{2}}} \right)}^{6}} \\  & {{=}^{6}}{{C}_{r}}{{(2x)}^{6-r}}{{\left( -\frac{3}{{{x}^{2}}} \right)}^{r}} \\  & {{=}^{6}}{{C}_{r}}{{2}^{6-r}}{{x}^{6-r}}{{\left( -1 \right)}^{r}}{{(3{{x}^{-2}})}^{r}} \\  & {{=}^{6}}{{C}_{r}}{{2}^{6-r}}{{x}^{6-r}}{{\left( -1 \right)}^{r}}({{3}^{r}}){{({{x}^{-2}})}^{r}} \\  & {{=}^{6}}{{C}_{r}}{{2}^{6-r}}{{\left( -1 \right)}^{r}}({{3}^{r}}){{x}^{6-r}}({{x}^{-2r}}) \\  & {{=}^{6}}{{C}_{r}}{{2}^{6-r}}{{\left( -1 \right)}^{r}}({{3}^{r}}){{x}^{6-3r}} \\  & \text{The term independent of }x\text{ will be when }{{x}^{0}} \\  & \therefore {{x}^{6-3r}}={{x}^{0}} \\  & 6-3r=0 \\  & r=2 \\  & \text{The term independent of }x\text{ will}{{\text{ }}^{6}}{{C}_{2}}{{2}^{6-2}}{{\left( -1 \right)}^{2}}({{3}^{2}}){{x}^{6-3(2)}} \\  & =15\times 16\times 9=2160 \\ \end{align}$ 

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