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Q.
The term independent of x in the expansion of $\left( 2x + \frac{1}{3x^2} \right)^9$ is
Binomial Theorem
Solution:
Suppose (r + 1)th term is independent of x.
We have
$T_{r+1} = ^{9} C_{r} \left(2x\right)^{9-r} \left(\frac{1}{3x^{2}}\right)^{r} $
$= ^{9}C_{r} 2^{9-r} \frac{1}{3^{r}} . x^{9-3r} $
This term is independent of x if 9 - 3r = 0
i.e., r = 3.
Thus, 4th term is independent of x.