Find the $4^{t h}$ term in the expansion of $(x - 2 y)^{12}$ .

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Solution

Here general term in the expansion of $(x - 2 y)^{12}$ is

$T_{r + 1} = 12_{C_{r}} (x)^{12 - r} . (- 2 y)^{r}$

$= (- 1)^{r} 12_{C_{r}} 2^{r} . x^{12 - r} . y^{r}$

Putting r=3,

$∴ T_{4} = (- 1)^{3} 12_{C_{3}} 2^{3} x^{12 - 3} . y^{3}$

$= - 12_{C_{3}} . 8 x^{9} y^{3}$

$= - 220 \times 8 x^{9} y^{3} = - 1760 x^{9} y^{3}$

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