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

Question

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

I have tried finding out the general term of the whole expression.

Then, I get the result as such $$[(-1)^r {3 \choose r}(x^{3-r})][{6 \choose r} (2x)^{6-r}(\frac{1}{x})^r]$$ But after solving for I get $r = 11/3$

I think I am doing it wrong by just multiplying the general term. I don't know how else to continue. Please help.

Thank You!

Edit : There was a typo in the question

Answer 1

You have\begin{align}\left(\frac1x+2x\right)^6&=\sum_{k=0}^6\binom6k\frac1{x^k}(2x)^{6-k}\\&=\sum_{k=0}^6\binom 6kx^{6-2k}2^{6-k}.\end{align}Here, the numbers that appear as exponents of $x$ belong to $\{-6,-4,-2,\ldots,6\}$. So, and since $(x-1)^3=x^3-3x^2+3x-1$ the coefficient of $x^{-2}$ in $(x-1)^3\left(\frac1x+2x\right)^6$ is the sum of these numbers:

• $-3\times\binom65\times2=-36$;
• $-\binom64\times2^2=-60$.

This sum is $-96$.

Answer 2

We have \begin{align}(x-1)^3(\frac1x+2x)^6&=\left(\frac1x\right)^6(x-1)^3(1+2x^2)^6\\ &=\frac1{x^6}\left(\sum_{r=0}^3\binom3rx^{3-r}(-1)^r\right)\left(\sum_{s=0}^6\binom6s(2x^2)^s\right)\\ \end{align}

In order to get an exponent of $-2$, we can ignore the $\dfrac1{x^6}$ and look for an exponent of $4$ instead. Therefore, we have $r=1$ and $s=1$, OR $r=3$ and $s=2$.

As such, we have $$\binom31(-1)^1\binom61(2)^1+\binom33(-1)^3\binom62(2)^2=\boxed{-96}.$$

Checking with WolframAlpha confirms the answer.

Answer 3

You have two binomial powers, so you need two separate binomial expansions. \begin{align*} (x-1)^3\left(\frac{1}{x} + 2x\right)^6 &= \sum_{r=0}^3 \binom{3}{r}(-1)^r x^{3-r} \sum_{s=0}^6 \binom{6}{s}(2x)^{6-s}\left(\frac{1}{x}\right)^s \\&= \sum_{r=0}^3\sum_{s=0}^6\binom{3}{r}\binom{6}{s}(-1)^r 2^{6-s} x^{(3-r)+(6-s)-s} \end{align*} If you want the coefficient of $x^{-2}$, you need to find the pairs $(r,s)$ with $0 \leq r \leq 3$ and $0 \leq s \leq 6$ such that \begin{align*} (3-r)+(6-s) - s &= -2 \\ \iff 9-r-2s &= -2 \\ \iff r+2s &= 11 \end{align*}

Answer 4

Hint. Write the product as $$-x^{-6}(1-x)^3(1+2x^2)^6.$$ Then the general term is given by $$-x^{-6}{3\choose j}(-x)^j{6\choose k}(2x^2)^k,$$ with $0\le j\le 3,\,0\le k\le 6.$ To get the term in $x^{-2},$ you therefore need to have $-2=-6+j+2k,$ or $j+2k=4,$ whose solutions are given by $(j,k)=(0,2),\,(2,1).$ Thus you want to compute the coefficient $$(-1)^{j+1}2^k{3\choose j}{6\choose k}$$ for these two solutions and add them up.