Binomial Theorem: Term Independent of x (Binomial Expansion)

The general term in $(a + b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here $n = 9, a = \frac{3x^2}{2}, b = -\frac{1}{3x}$: $$T_{r+1} = \binom{9}{r} \left( \frac{3x^2}{2} \right)^{9-r} \left( -\frac{1}{3x} \right)^r$$ $$= \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{2(9-r) - r} = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} (-1)^r \left( \frac{1}{3} \right)^r x^{18 - 3r}$$

For the term to be independent of $x$: $$18 - 3r = 0 \implies 3r = 18 \implies r = 6$$ Thus, the 7th term ($T_7$) is independent of $x$.

$$T_7 = \binom{9}{6} \left( \frac{3}{2} \right)^3 (-1)^6 \left( \frac{1}{3} \right)^6 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \frac{27}{8} \times 1 \times \frac{1}{729}$$ $$= 84 \times \frac{27}{8} \times \frac{1}{729} = 84 \times \frac{1}{8 \times 27} = \frac{84}{216} = \frac{7}{18}$$