A question involving the expansion of $(x+2)^n$

Question

I have the following question:

In the expansion of $(x+2)^n$, $n$ is a positive integer greater than $2$. Given that the ratio between the coefficient of the $x^3$ term and the coefficient of the $x$ term is $7:4$, find $n$.

I know I have to use binomial theorem, but I'm not too sure how to go from there. I get that the coefficient of the $x^3$ term is $\binom{n}32^{n-3}$ and the coefficient of the $x$ term to be $n*2^{n-1}$. So I get the equation:$$\frac{\binom{n}32^{n-3}}{n*2^{n-1}}=\frac74$$Am I on the right track? I'm also not too sure how to solve this equation, so could someone help me with that?

Answer 1

${n \choose 3} = \frac{n!}{(n-3)! \ 3!} = \frac{1}{6}n(n-1)(n-2)$. Therefore, after cancelling you have:

$$\frac{(1/6)(n-1)(n-2)}{4} = \frac{7}{4}.$$

Answer 2

What you have done is correct. To complete just write down $\binom {n} {3}$ as $\frac {n(n-1)(n-2)} 6$ to get the equation $n^{2}-3n-40=0$ which gives $n=8$.

Answer 3

As the expansion contains powers of $2$ and the terms in question are spaced by two positions, you need to find two binomial coefficients in the ratio $7$.

As the ratios go increasing, this is the only solution.