Question

Need help with this ASAP​

Answer 1

Given:

$(3x-2)^9$

To find:

The coefficient of $x^6$ in the expansion of $(3x-2)^9$.

Solution:

We know that, in the expansion of $(x+y)^n$,

$T_{r+1}=^nC_rx^{n-r}y^r$       ...(i)

$(3x-2)^9$ has only one x and in $x^6$, the power of x is 6. It is possible if

$n-r=6$

$9-r=6$

$9-6=r$

$3=r$

Putting n=9, r=3, x=3x and y=-2 in (i), we get

$T_{3+1}=^9C_3(3x)^{9-3}(-2)^3$

$T_{4}=\dfrac{9!}{3!(9-3)!}\times (3x)^{6}(-8)$

$T_{4}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times (9-3)!}\times (729x^{6})(-8)$

$T_{4}=84\times (729)\times (-8)\times x^{6}$

$T_{4}=-489888x^6$

Therefore, the coefficient of $x^6$ in the expansion of $(3x-2)^9$ is -489888.