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

Open in App

Solution

It is known that $(r + 1)$ th term $(T_{r + 1})$ in the binomial expansion of
$(a + b)^{n}$ is given by $T_{r + 1} =^{n} C_{r} b^{n - r} b^{r}$
Thus the 4th term in the expansion of $(x + 2 y)^{12}$ is
$T_{4} = T_{3 + 1} =^{12} C_{3} (x)^{12 - 3} (- 2 y)^{3} = (- 1)^{3} \frac{12!}{3! 9!} . x^{9} . (2)^{3} . y^{3}$
$= - \frac{12.11.10}{3.2} . (2)^{3} x^{9} y^{3} = - 1760 x^{9} y^{3}$

Suggest Corrections

Similar questions

{(x + \frac{2}{x})}^{9}

{(x + y)}^{n}

4 t h

13 t h

^{'} n^{'}

{(x + y)}^{n}

240, 720, 1080

x, y, n

4^{t h}

{(p x + \frac{1}{x})}^{n}

x

n

p

4^{t h}

5 / 2

Join BYJU'S Learning Program