# Find the quotient of terms containing x^28 and x^-4 in the expansion of (2x^6-3/x^2)^10?

Oct 1, 2017

$103680$ for ${x}^{28}$ and $1180980$ for ${x}^{- 4}$

#### Explanation:

Expansion of ${\left(a + b\right)}^{10}$ will have $11$ terms and its ${r}^{t h}$ term is given by

${C}_{r}^{10} {a}^{r} \cdot {b}^{10 - r}$

Here $a = 2 {x}^{6}$ and $b = - \frac{3}{x} ^ 2$

Hence ${r}^{t h}$ term is ${C}_{r}^{10} {\left(2 {x}^{6}\right)}^{r} \cdot {\left(- \frac{3}{x} ^ 2\right)}^{10 - r}$

= ${C}_{r}^{10} {2}^{r} \cdot {x}^{6 r} \cdot {\left(- 3\right)}^{10 - r} {\left({x}^{- 2}\right)}^{10 - r}$

= ${C}_{r}^{10} {2}^{r} \cdot {x}^{6 r} \cdot {\left(- 3\right)}^{10 - r} \left({x}^{- 20 + 2 r}\right)$

= ${C}_{r}^{10} {2}^{r} \cdot {\left(- 3\right)}^{10 - r} \cdot {x}^{6 r - 20 + 2 r}$

If we have ${x}^{28}$. $6 r - 20 + 2 r = 28$ i.e. $8 r = 48$ or $r = 6$ and corresponding term is

${C}_{8}^{10} {2}^{8} \cdot {\left(- 3\right)}^{2} \cdot {x}^{28} = \frac{10 \times 9}{1 \times 2} \cdot 256 \cdot 9 {x}^{28} = 103680 {x}^{28}$

and if we have ${x}^{- 4}$. $6 r - 20 + 2 r = - 4$ i.e. $8 r = 16$ or $r = 2$ and corresponding term is

${C}_{2}^{10} {2}^{2} \cdot {\left(- 3\right)}^{8} \cdot {x}^{- 4} = \frac{10 \times 9}{1 \times 2} \cdot 4 \cdot 6561 {x}^{- 4} = 1180980 {x}^{- 4}$