If the term independent of x in the expansion of (32x^2-13x)^9 is k, then 18k is equal to:

Question

If the term independent of $x$ in the expansion of $\left(\dfrac{3}{2}x^2-\dfrac{1}{3x}\right)^9$ is $k$, then $18k$ is equal to:

Accepted Answer

$$T_{r+1}=\binom9r\left(\frac32x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r=\binom9r\left(\frac32\right)^{9-r}\left(-\frac13\right)^r\,x^{18-3r}$$

Independent of $x$: $18-3r=0\Rightarrow r=6$

$$T_7=\binom96\left(\frac32\right)^3\left(-\frac13\right)^6=\binom93\cdot\frac{1}{6^3}=84\times\frac{1}{216}=\frac{7}{18}$$

$$18k=18\times\frac{7}{18}=7$$

Solutions · 1