Find the coefficient of x4 in the expansion of (3x - 1/x)6

2. $\displaystyle\left(3x-\frac{1}{x}\right)^6$

The general term is $T_{k+1}=(-1)^kC_6^k(3x)^{6-k}\left(\frac{1}{x}\right)^k=(-1)^kC_6^k3^{6-k}x^{6-2k}$

$x^{6-2k}=x^4\Rightarrow 6-2k=4\Rightarrow k=1$
So, the second term is $T_2=-6\cdot 3^5\cdot x^4=-1458x^4$

3. Note that there is certainly a MUCH shorter way doing this. However, for good pedagogy, I worked this expansion out completely for you (so you understand each step). I will assume you know how to compute binomial coefficients (i.e. 6 choose 4).

Find the coefficient of x4 in the expansion of (3x - 1/x)6

We want to expand $(3x - x^{-1})^6$.

The Binomial Theorem states:

$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

We let:

$a = 3x$

$b = -x^{-1}$

$n = 6$

So, $(3x - x^{-1})^6 = \sum_{k=0}^6 \binom{6}{k} (3x)^{6-k} (-x^{-1})^k$

$= \binom{6}{0}(3x)^6 (-x^{-1})^0 + \binom{6}{1}(3x)^5 (-x^{-1})^1 + \binom{6}{2}(3x)^4 (-x^{-1})^2 + \binom{6}{3}(3x)^3$ $(-x^{-1})^3 + \binom{6}{4}(3x)^2 (-x^{-1})^4 + \binom{6}{5}(3x)^1 (-x^{-1})^5 + \binom{6}{6}(3x)^0 (-x^{-1})^6$

$= \binom{6}{0}729x^6 - \binom{6}{1}243x^5 x^{-1} + \binom{6}{2}81x^4 x^{-2} - \binom{6}{3}27x^3$ $x^{-3} + \binom{6}{4}9x^2 x^{-4} - \binom{6}{5}3x x^{-5} + \binom{6}{6} 1^{-6}$

$= \binom{6}{0}729x^6 - \binom{6}{1}243x^4 + \binom{6}{2}81x^2 - \binom{6}{3}27$ $+ \binom{6}{4}9x^{-2} - \binom{6}{5}3x^{-4} + \binom{6}{6} 1^{-6}$

$= 1(729x^6) - 6(243x^4) + 15(81x^2) - 20(27) + 15(9x^{-2}) - 6(3x^{-4}) + 1(1^{-6})$

$= 729x^6 - 1458x^4 + 1215x^2 - 540 + \frac{135}{x^2} - \frac{18}{x^4} + \frac{1}{x^6}$

So, what does the coefficient of $x^4$ look like to you?

-Andy