Finding the exponent $n$ in a binomial expansion given that the coefficient of $x^3$ is 4 times greater than the coefficient of $x^2$

Question

Consider the expansion of $(2+x)^n$ where $n >=3$ and $n \epsilon Z$. The coefficient of $x^3$is four times the coefficient of $^2$. Find the value of $n$.

Answer 1

The coefficient of $x^3$ is

$$2^{n-3}{n\choose 3}$$

And the coefficient of $x^2$ is

$$2^{n-2}{n\choose 2}$$

So we have

$${2^{n-3}\over 2^{n-2}}{2!(n-2)!\over n!}{n!\over 3!(n-3)!}=4$$

This simplifies to $n-2=24$ leading to $n=26$