Which term is independent of x in the expansion of

Question:

Which term is independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$

Solution:

To find: the term independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$

Formula Used:

A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $\left(x-\frac{1}{3 x^{2}}\right)^{9}$ we get

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

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

$T_{r+1}={ }^{9} C_{r} \times(-1) \times 3 x^{9-3 r}$

For finding the term which is independent of x,

$9-3 r=0$

r=3

Thus, the term which would be independent of $x$ is $T_{4}$

Thus, the term independent of $x$ in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ is $T_{4}$ i.e $4^{\text {th }}$ term

$T_{6}={ }^{10} C_{5} \times\left(\frac{2 x^{2}}{3}\right)^{5} \times\left(\frac{3}{2 x^{2}}\right)^{5}$

$T_{6}={ }^{10} C_{5}$

$T_{6}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2}$

$T_{6}=252$

Thus, the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is 252 .