Question

Find the term independent of x in the expansion of ${(\frac{3x^2}{2}-\frac{1}{3x})}^{15}$

Answer 1

Let $T_{r+1}$ be independent of x.
Here, $T_{r+1}=(-1{)}^r{\times }^{15}{C}_r\times {\left(\frac{3x^2}{2}\right)}^{(15-r)}\times {\left(\frac{1}{3x}\right)}^r\Rightarrow T_{r+1}=(-1{)}^r{\times }^{15}{C}_r\times {30}^{(15-2r)}\times 2^{(r-15)}\times x^{(30-3r)}$
Now, $T_{r+1}$ will be independent of x only when the power of x in it is 0

$\therefore 30-3r\Rightarrow 3r=30\Rightarrow r=10\Rightarrow r+1=11$
Thus $T_{11}$ is independent of x.
Now, $T_{11}=T_{(10+1)}=(-1{)}^{10}{\times }^{15}{C}_{10}\times 3^{(15-20)}\times 2^{(10-15)}\times x^0{=}^{15}{C}_5\times 3^{-5}\times 2^{-5}=\frac{{}^{15}{C}_5}{6^5}=(\frac{15\times 14\times 13\times 12\times 11}{5\times 4\times 3\times 2\times 1}\times \frac{1}{36\times 36\times 6})=\frac{1001}{2592}$
Hence the term independent of x in the given expansion is $\frac{1001}{2592}$