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?

Ooops ... I mean't\begin{eqnarray} \frac{1}{(1-x)^{n+1}} = \sum_{k=0}^{\infty} \binom{n+k}{k}x^k. \end{eqnarray} — Donald Splutterwit, Nov 15 '19 at 21:36
The coefficients are given recursively by $a_0=1$, $a_{n+1}=-a_n+(-1)^{n+1}n$ for $n>0$. Can you take it from here? Alternatively, you can try manipulating $\frac1{1-x}$ which has a well-known power series expansion for $|x|<1$. — user170231, Nov 15 '19 at 21:39
"It is not quite an arithmetic series"... Well, it is an arithmetic series with alternating signs. I don't know why my question as been downvoted. Check it by yourself! — the_candyman, Nov 15 '19 at 21:50

score 2 · Answer 1 · answered Nov 15 '19 at 21:46

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 $$ \frac {2}{(1+x)^3} = 2-6x+12x^2-....$$ $$ \frac {1}{(1+x)^3} = 1-3x+6x^2-....$$

answered Nov 15 '19 at 21:46

Mohammad Riazi-Kermani

69,891

+1 for the smart usage of derivatives – the_candyman Nov 15 '19 at 21:56

score 2 · Answer 2 · answered Nov 15 '19 at 21:46

2

If you know calculus, then

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

answered Nov 15 '19 at 21:46

Doug M

58,694

the_candyman · Accepted Answer · 2019-11-15T22:12:53.823

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

edited Nov 15 '19 at 22:12

answered Nov 15 '19 at 21:24

the_candyman

14,234
4
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So $\binom{n+2}{n}=\dfrac{n(n-1)(n-2)(n-n+1)}{2!}$
Is this correct?
– James Warthington Nov 15 '19 at 21:59
It is $-3$, so it should be $\binom{n-4}{n}$, no? – James Warthington Nov 15 '19 at 22:12
@JamesWarthington, sorry, I misreported it. – the_candyman Nov 15 '19 at 22:13
so it is $\binom{n-4}{n}$, right? – James Warthington Nov 15 '19 at 22:59
No, $\alpha = 3$, not $\alpha = -3$... – the_candyman Nov 16 '19 at 09:00
Moreover, notice that: $${{n + 2}\choose{n}} = \frac{(n+2)!}{n! (n+2-n)!} = \frac{(n+2)!}{n! 2!} = \frac{(n+2)(n+1)n!}{n! 2!} = \frac{(n+2)(n+1)}{2}.$$ – the_candyman Nov 16 '19 at 09:14

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

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