General formula for the power series of $\dfrac{1}{(1+x)^3}$ I wish to find the general formula for the following power series:
$\dfrac{1}{(1+x)^3}=1-3x+6x^2-10x^3+15x^4-21x^5+28x^6...$
The difference between the first and second term is $2$
The difference between the second and third term is $3$
The difference between the third and fourth term is $4$
The difference between the fourth and fifth term is $5$
and so on.
How do you find the general formula for this series? It is not quite an arithmetic series, and definitely not a geometric series, so what should I do?
 A: $$\frac {1}{1+x} = 1-x+x^2-x^3+x^4-....$$
Differentiate and you get 
$$ \frac {-1}{(1+x)^2} = -1+2x-3x^2+4x^3-....$$
Differentiate again and you get 
$$ \frac {2}{(1+x)^3} = 2-6x+12x^2-....$$
$$ \frac {1}{(1+x)^3} = 1-3x+6x^2-....$$ 
A: If you know calculus, then
$\frac {1}{(1+x)^3} = \frac {d^2}{dx^2}\frac {1}{2(1+x)} = \frac {d^2}{dx^2} \sum_\limits{n=0}^\infty \frac {(-1)^nx^n}{2} = \sum_\limits{n=0}^\infty \frac {(-1)^n(n+1)(n+2) x^n}{2}$
A: In general, evaluating the Taylor expansion of your function in a neighborhood of $0$ yields to
$$\frac{1}{(x+1)^3} = \sum_{n=0}^{+\infty}\frac{f^{(n)}(0)}{n!}x^n =\sum_{n=0}^{+\infty} a_n x^n,$$
where $a_n = \frac{f^{(n)}(0)}{n!}.$
In your case, it can be proven using binomial series that:
$$a_n = (-1)^{n} \frac{(n+1)(n+2)}{2}.$$
Indeed, it is well-known that:
$$(1-z)^{-\alpha} = \sum_{n=0}^{+\infty}{{n + \alpha - 1}\choose{n}}z^n.$$
Therefore:
$$\begin{align*}(1-(-x))^{-3} &= & \sum_{n=0}^{+\infty}{{n + 2}\choose{n}}(-x)^n = \\
& = & \sum_{n=0}^{+\infty}\frac{(n+2)!}{n!2!}(-1)^n x^n = \\
& = & \sum_{n=0}^{+\infty}\left[\frac{(n+1)(n+2)}{2}(-1)^n\right]x^n.\end{align*}$$
