You are given the following expression:
$$(ax + by)^{n} = -15120x^{4}y^{3}$$
Determine the constants $a$, $b$ and $n$.
My attempt to solve this problem is by trying to use the binomial theorem backwards.
The binomial theorem: $(ax + by)^{n} = \displaystyle\sum_{k = 0}^{n} {\binom{n}{k}(ax)^{n-k}(by)^{k}}$
And so if you compare the expressions, you get:
$$\binom{n}{k}(a)^{n-k}(b)^{k} = -15120$$
$n – k = 4$ and $k = 3$ so that $n = 7$
Here is where I get stuck, because now I have:
$$\binom{7}{3}(a)^{4}(b)^{3} = -15120$$
Two unknowns… How to solve it? Am I even doing it correctly?
Best Answer
I think, that the exercise is
In this case you know, that the 5th summand is ${7 \choose 4} \cdot (ax)^4\cdot (by)^3=-15120x^{4}y^{3}$
$15120=2^4\cdot3^3\cdot5\cdot7$.
${7 \choose 4}=5\cdot 7$
Thus $b=-3$ and $a=2$