Find the term independent of x in the expansion of the following expressions:

Solution:

(i) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$

$T_{r+1}={ }^{9} C_{r}\left(\frac{3}{2} x^{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^{r}$

$=(-1)^{r}{ }^{9} C_{r} \cdot \frac{3^{9-2 r}}{2^{9-r}} \times x^{18-2 r-r}$

For this term to be independent of $x$, we must have

$18-3 r=0$

$\Rightarrow 3 r=18$

$\Rightarrow r=6$

Hence, the required term is the 7 th term.

Now, we have

${ }^{9} C_{6} \times \frac{3^{9-12}}{2^{9-6}}$

$=\frac{9 \times 8 \times 7}{3 \times 2} \times 3^{-3} \times 2^{-3}$

$=\frac{7}{18}$

(ii) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$

$T_{r+1}={ }^{9} C_{r}(2 x)^{9-r}\left(\frac{1}{3 x^{2}}\right)^{r}$

$={ }^{9} C_{r} \cdot \frac{2^{9-r}}{3^{r}} x^{9-r-2 r}$

For this term to be independent of $x$, we must have

$9-3 r=0$

$\Rightarrow r=3$

Hence, the required term is the 4 th term.

Now, we have

${ }^{9} C_{3} \frac{2^{6}}{3^{3}}$

$={ }^{9} C_{3} \times \frac{64}{27}$

(iii) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(2 x^{2}-\frac{3}{x^{3}}\right)^{25}$

$T_{r+1}={ }^{25} C_{r}\left(2 x^{2}\right)^{25-r}\left(\frac{-3}{x^{3}}\right)^{r}$

$=(-1)^{r}{ }^{25} C_{r} \times 2^{25-r} \times 3^{r} x^{50-2 r-3 r}$

For this term to be independent of $x$, we must have:

$50-5 r=0$

$\Rightarrow r=10$

Therefore, the required term is the 11 th term.

Now, we have

$(-1)^{10}{ }^{25} C_{10} \times 2^{25-10} \times 3^{10}$

$={ }^{25} C_{10}\left(2^{15} \times 3^{10}\right)$

(iv)  Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(3 x-\frac{2}{x^{2}}\right)^{15}$

$T_{r+1}={ }^{15} C_{r}(3 x)^{15-r}\left(\frac{-2}{x^{2}}\right)^{r}$

$=(-1)^{r}{ }^{15} C_{r} \times 3^{15-r} \times 2^{r} x^{15-r-2 r}$

For this term to be independent of $x$, we must have

$15-3 r=0$

$\Rightarrow r=5$

Hence, the required term is the 6 th term.

Now, we have :

$(-1)^{5}{ }^{15} C_{5} \cdot 3^{15-5} \cdot 2^{5}$

$=-3003 \times 3^{10} \times 2^{5}$

(v) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$

$T_{r+1}={ }^{10} C_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r}$

$={ }^{10} C_{r} \cdot \frac{3^{r-\frac{10-r}{2}}}{2^{r}} x^{\frac{10-r}{2}-2 r}$

For this term to be independent of $x$, we must have

$\frac{10-r}{2}-2 r=0$

$\Rightarrow 10-5 r=0$

$\Rightarrow r=2$

Hence, the required term is the 3 rd term.

Now, we have

$=\frac{10 \times 9}{2 \times 4 \times 9}$

$=\frac{5}{4}$

(vi)  Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(x-1 / x^{2}\right)^{3 n}$

$T_{r+1}={ }^{3 n} C_{r} x^{3 n-r}\left(\frac{-1}{x^{2}}\right)^{r}$

$=(-1)^{r}{ }^{3 n} C_{r} x^{3 n-r-2 r}$

For this term to be independent of $x$, we must have

$3 n-3 r=0$

$\Rightarrow r=n$

Hence, the required term is the $(n+1)$ th term.

Now, we have

$(-1)^{n}{ }^{3 n} C_{n}$

(vii)  Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(\frac{1}{2} x^{1 / 3}+x^{-1 / 5}\right)^{8}$

$T_{r+1}={ }^{8} C_{r}\left(\frac{1}{2} x^{1 / 3}\right)^{8-r}\left(x^{-1 / 5}\right)^{r}$

$={ }^{8} C_{r} \cdot \frac{1}{2^{8-r}} x^{\frac{8-r}{3}-\frac{r}{5}}$

For this term to be independent of $\mathrm{x}$, we must have

$\frac{8-r}{3}-\frac{r}{5}=0$

$\Rightarrow 40-5 r-3 r=0$

$\Rightarrow 8 r=40$

$\Rightarrow r=5$

Hence, the required term is the 6 th term.

Now, we have :

${ }^{8} C_{5} \times \frac{1}{2^{8-5}}$

$=\frac{8 \times 7 \times 6}{3 \times 2 \times 8}=7$

(ix) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}$

$T_{r+1}={ }^{18} C_{r}\left(x^{1 / 3}\right)^{18-r}\left(\frac{1}{2 x^{1 / 3}}\right)^{r}$

$={ }^{18} C_{r} \times \frac{1}{2^{r}} x^{\frac{18-r}{3}-\frac{r}{3}}$

For this term to be independent of $r$, we must have

$\frac{18-r}{3}-\frac{r}{3}=0$

$\Rightarrow 18-2 r=0$

$\Rightarrow r=9$

The term is

${ }^{18} C_{9} \times \frac{1}{2^{9}}$

(x) Suppose the (r + 1)th term in the given expression is independent of x.

Now,

$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$

$T_{r+1}={ }^{6} C_{r}\left(\frac{3}{2} x^{2}\right)^{6-r}\left(\frac{-1}{3 x}\right)^{r}$

$=(-1)^{r^{6}} C_{r} \times \frac{3^{6-r-r}}{2^{6-r}} x^{12-2 r-r}$

For this term to be independent of $x$, we must have

$12-3 r=0$

$\Rightarrow r=4$

Hence, the required term is the 4 th term.

${ }^{6} C_{4} \times \frac{3^{6-4-4}}{2^{6-4}}$

$=\frac{6 \times 5}{2 \times 1 \times 4 \times 9}$

$=\frac{5}{12}$