Question

find the coefficient of x in the expansion of
$( \frac{x}{3} + \frac{9}{ {x}^{2} } ) ^{7}$
​

Answer 1

By the binomial theorem,

$\left(\dfrac x3+\dfrac9{x^2}\right)^7=\displaystyle\sum_{n=0}^7\binom7n\left(\frac x3\right)^{7-n}\left(\frac9{x^2}\right)^7=\sum_{n=0}^7\binom7n\left(\frac3x\right)^{3n-7}$

We get the $x$ term when 3n - 7 = -1, or n = 2, which has coefficient

$\dbinom723^{3\cdot2-7}=\boxed{7}$