How to simplify the the binomial coefficient in the binomial series? Use binomial series to expand the function $\frac{5}{(6+x)^3}$ as a power series.
I understand the process to get the following summation:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {-3 \choose n} (\frac{x}{6})^n $
However, I am stuck on seeing what's going on with ${-3 \choose n}$.
From the Stewart Calculus textbook, it says that ${k \choose n} = \frac{k(k-1)(k-2)...(k-n+1)}{n!}$.
By applying that, I would get:
${-3 \choose n}=\frac {(-3)(-4)(-5)...[-(n+2)]}{n!}$.
I think the next step would be to extract out the negative, so I will get $(-1)^n$.
The summation would then be:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {\frac {(-1)^n(3)(4)(5)...[(n+2)]}{n!}} (\frac{x}{6})^n $
The solution to this problem is $\frac{5}{2}\sum_{n=0}^{\infty} {\frac {(-1)^n(n+1)(n+2)x^n}{6^{n+3}}}$
I am not sure what has happened to the factorial. Was it cancelled out due to the (3)(4)(5)... in the numerator? Where did (n+1) come from?

$$\frac {(3)(4)(5)\cdots(n+2)}{n!} = \frac {(3)(4)(5)\cdots(n)(n+1)(n+2)}{(1)(2)(3)(4)(5)\cdots(n)} = \frac {(n+1)(n+2)}{(1)(2)}$$
