# How to find coefficient of multiple binomial expansions quickly?

Find the coefficient of $(x^{-2})$ in the expansion of $(x-1)^3(\frac1{x}+2x)^6$

How can this be done quickly without expanding the two brackets?

I tried writing the general formula in terms of $r$ for both brackets and equated $-2$ to the sum of the powers of $x$ in terms of $r$. $${3 \choose r} (x^{3-r}) (-1)^r {6 \choose r} \left(\frac1{x}^6-r\right)(2x)^r$$

I understand that there are multiple ways of getting $x^{-2}$, but shouldn't this be accounted for in the general formula method? (multiple $x^{-2}$ terms can be simplified to one number which should obey the general term formula).

To convert to a more familiar form, multiply and divide by $x^6$:

$$(x-1)^3 (1+2x^2)^6 x^{-6}$$

Now you're looking for the coefficient of $x^4$ in $(x-1)^3 (1+2x^2)^6$. You can get $x^4$ out of this product either by multiplying $x^2$ with $x^2$ or $1$ with $x^4$. Each of those coefficients should be easy to calculate, then you just add them.

Here is a variation which goes consecutively through the factors expanding one by one up to terms which are needed ignoring other terms. We also start with the factors with greatest exponents which keep the number of terms which are to expand small.

It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain \begin{align*} \color{blue}{[x^{-2}]}&\color{blue}{(x-1)^3\left(\frac{1}{x}+2x\right)^6}\\ &=[x^4](1+2x^2)^6(x-1)^3\tag{1}\\ &=[x^4]\left(1+\binom{6}{1}2x^2+\binom{6}{2}4x^4\right)(x-1)^3\tag{2}\\ &=\left([x^4]+12[x^2]+60[x^0]\right)(x-1)^3\tag{3}\\ &=12\binom{3}{2}(-1)+60\binom{3}{0}(-1)^3\tag{4}\\ &\,\,\color{blue}{=-96} \end{align*}

Comment:

• In (1) we factor out $\frac{1}{x^6}$ and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (2) we expand the $(1+2x^2)^6$ up to terms which contribute to $[x^4]$. Other terms can be ignored.

• In (3) use the linearity of the coefficient of operator.

• In (4) we select the coefficients of $(x-1)^3$ accordingly.

• @RoroGVA: You're welcome. – Markus Scheuer Mar 28 '18 at 8:13