Binomial theorem to find $n$

Question

If the coefficient of $x^3$ in the expansion $(3x^2-2/x)^n$ is $-54$, find the value of $n$.

When I tried to do it using the normal expansion $nCr$, I ended up with 2 simultaneous equations (A&B) which cannot be solved.

$2n=3r$----A $\qquad nCr*3^{(n+1-r)}*(-2)^{(r-1)}=-54$ ----B

Please help :)

Answer 1

HINT Using the Binomial Theorem, you should get $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} $$ In your case, $a = 3x^2$ and $b = -2/x$. Can you

1. plug in and simplify the terms
2. find the term with $x^3$ -- what value(s) of $k$ would it correspond to?
3. compute its coefficient?

Answer 2

Using the binomial formula: $$\Bigl(3x^2-\frac2x\Bigr)^n=\sum_{r=0}^n(-1)^{n-r}\binom nr 3^rx^{2r} \frac{2^{n-r}}{x^{n-r}}=\sum_{r=0}^n(-1)^{n-r}\binom nr 3^r2^{n-r} x^{3r-n}$$ This formula shows the coefficient of $x^3$ is $-54$ if and only if $$(\mathrm i)\;3r-n=3,\qquad (\mathrm{ii})\;n-r \text{ is odd },\qquad (\mathrm{iii})\;\binom nr 3^r2^{n-r}=54=2\cdot 3^3 $$ Equation (iii) shows $\dbinom nr2^{n-r}$ has $2$-valuation equal to $1$, hence $n-r\le 1$. Observe $n-r$ can't be $0$ since then $\dbinom nr=1$ and the l.h.s. of (iii) is equal to $3^r$. So, necessarily $n-r=1$, i.e. $r=n-1$, and eq. (i) becomes $$3(n-1)-n=3\iff 2n=6.$$ Thus we have $n=3,\;r=2$ (note condition (ii) is satisfied).