If the 4th term in the expansion of $\left( \frac{2}{3}x - \frac{3}{2x} \right)^{n}$ is independent of $x$, then $n$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
Given expansion is $\left( \frac{2}{3}x - \frac{3}{2x} ight)^{n}$
$	herefore\ T_{4} = \ ^{n}C_{3}\left( \frac{2}{3}x ight)^{n - 3}\left( - \frac{3}{2x} ight)^{3}$
$= \ ^{n}C_{3}\left( \frac{2}{3} ight)^{n - 6}x^{n - 6}{( - 1)}^{3}$
Since, it is independent of $x$
$	herefore\ n - 6 = 0\  \Rightarrow \ n = 6$

Q. If the 4th term in the expansion of $\left( \frac{2}{3}x - \frac{3}{2x} \right)^{n}$ is independent of $x$, then $n$ is equal to

Explanation

Reviewed by Dr. Abubacker

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