General formula for the power series of $\dfrac{1}{(1+x)^3}$

Question

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?

Answer 1

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: $$\begin{align*}(1-(-x))^{-3} &= & \sum_{n=0}^{+\infty}{{n + 2}\choose{n}}(-x)^n = \\ & = & \sum_{n=0}^{+\infty}\frac{(n+2)!}{n!2!}(-1)^n x^n = \\ & = & \sum_{n=0}^{+\infty}\left[\frac{(n+1)(n+2)}{2}(-1)^n\right]x^n.\end{align*}$$

Answer 2

$$\frac {1}{1+x} = 1-x+x^2-x^3+x^4-....$$

Differentiate and you get

$$ \frac {-1}{(1+x)^2} = -1+2x-3x^2+4x^3-....$$

Differentiate again and you get $$ \frac {2}{(1+x)^3} = 2-6x+12x^2-....$$ $$ \frac {1}{(1+x)^3} = 1-3x+6x^2-....$$

Answer 3

If you know calculus, then

$\frac {1}{(1+x)^3} = \frac {d^2}{dx^2}\frac {1}{2(1+x)} = \frac {d^2}{dx^2} \sum_\limits{n=0}^\infty \frac {(-1)^nx^n}{2} = \sum_\limits{n=0}^\infty \frac {(-1)^n(n+1)(n+2) x^n}{2}$