I was asked to find the first $3$ terms of the expansion $\left(3-\frac1{9x}\right)^5$ and was further asked to find the term independent of x in the expansion of $\left(3-\frac1{9x}\right)^5(2+9x)^2$.
So far, I expanded those $2$ and got: $$\left(243-\frac{45}x+\frac{10}{3x^2}\right)(4+36x+81x^2)$$ And... I am stuck...
ANS is -378
On
HINT
Just multiply the "inverse" terms; the one with $x^0$ with the other one, the one with $x^1$ with the one with $x^{-1}$ and finally the one with $x^2$ with the one with $x^{-2}$ and you will get your solution since all other terms you can produce like this will contain at least an $x$ or an $x^{-1}$.
On
HINT. The Binomial Theorem says that $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^{n}$.
The first is ambiguous depending on how one decides to expand. One could do in order of power or the expansion order, but we'll assume its expansion order so that the first three terms are $\binom{5}{k} a^{5-k} b^n$ for $k=0,1,2$, where $a=3$, $b= -\frac{1}{9x}$.
For the second part, the only way to get a constant out of the product is the following:
$1)$ Constant $\times$ Constant: What is the constant term of the expansions $(3-1/(9x))^5$ and $(2+9x)^2?$
$2)$ $\frac{1}{x} \times x$: What is the $\frac{1}{x}$ term in the expansion of $(3-1/(9x))^5$ and what is the $x$ term for $(2+9x)^2?$
$3)$ $\frac{1}{x^2} \times x^2$: What is the $\frac{1}{x^2}$ term in the expansion of $(3-1/(9x))^5$ and what is the $x^2$ term for $(2+9x)^2?$
This will give you the three constants that come out in the expansion of the product. Adding them will give you the constant term, the only term independent of $x$ in the expansion.
Just notice that $$243\times 4 - \frac{45}{x}\times 36x + \frac{10}{3x^2}\times 81x^2$$ is the term independent of $x$.