Originally Posted by

**o_O** Binomial series are given by: $\displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$

for any $\displaystyle k \in \mathbb{R}$ and if $\displaystyle |y| < 1$. Here, $\displaystyle y = \tfrac{x}{10}$ and $\displaystyle k = -4$ .

So: $\displaystyle g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$

$\displaystyle g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $\displaystyle + \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$

_______________

Let's look at the general term: $\displaystyle \frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$

Just looking at the expanded series, we see that with even $\displaystyle n$, the coefficient is positive and with odd $\displaystyle n$, the coefficient is negative.

So, if we factor out the negative signs: $\displaystyle \frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$

etc.