Coefficients and Expansions So I was just hoping for a look over my work to check if what I am doing is right because I'm not so sure:
Find the coefficient of: f$$ x^6$$ with the equation $$(3x-\frac{(1)}{x^2})^{12}$$
I have: $\sum_{i=0}^{24}  $$\binom{12}{k}(3x)^{12-k}(-x^{-2})^{k}$ and simplified it. In the end, I got k = 2.
I got $$\binom{12}{2}(3)^{10}$$ as my coefficient, is that correct?
 A: $$
\left(3x-\frac{1}{x^2}\right)^{12} = \sum_{i=0}^{12} \binom{12}{k}(3x)^{12-k}(-x^{-2})^{k}
$$
$$
\binom{12}{k}(3x)^{12-k}(-x^{-2})^{k} = \binom{12}{k}(3)^{12-k}(x)^{12-k}(-1)^k x^{-2k} = \binom{12}{k}(3)^{12-k}(-1)^k(x)^{12-3k}
$$
$12 - 3k = 6 \rightarrow k = 2$
$$
\binom{12}{2}3^{10}x^{6}
$$
A: Looks good.
In your approach you had to solve $12-k+-2k=6$ to get $k=2$. Sometimes this isn't so obvious/easy so an alternative is to write it as:
$$\left(\frac{3x^3-1}{x^2}\right)^{12}=\frac{\left(x^3-1\right)^{12}}{x^{24}}$$
This changes the problem to finding the coefficient of $x^{30}$ in the numerator. It is then more obvious which term (and hence coefficient) you want So we need:
$${12\choose10}(3x^3)^{10}(-1)^2$$
So the coefficient is $${12\choose10}3^{10}$$
Note:${12\choose10}={12\choose2}$
A: You are right. We are looking for terms of the form $\binom {12}{A}3^Ax^A(-1)^Bx^{-2 B}$ in the binomial expansion, in which $A-2 B=6$ and $A+B=12$, which requires $A=10, B=2.$ So there is just one term, which is the one you got.
A: It is   convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ of a polynomial $P(x)$. We can write
\begin{align*}
[x^j]P(x)=[x^j]\sum_{k=0}^na_kx^k=a_j
\end{align*}

We obtain 
  \begin{align*}
[x^6]&\left(3x-\frac{1}{x^2}\right)^{12}\\
&=[x^6]\frac{1}{x^{24}}\left(3x^{3}-1\right)^{12}\tag{1}\\
&=[x^{30}]\sum_{k=0}^{12}\binom{12}{k}\left(3x^3\right)^k(-1)^{12-k}\tag{2}\\
&=[x^{30}]\sum_{k=0}^{12}\binom{12}{k}3^{k}x^{3k}(-1)^{12-k}\tag{3}\\
&=\binom{12}{10}3^{10}
\end{align*}

Comment:


*

*In (1) we extract $\frac{1}{x^2}$

*In (2) we expand the binomial and apply the rule $[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$

*In (3) we do a small rearrangement and see that $k=10$ gives the coefficient of $x^{30}$ which we want to extract.
