Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I find the coefficient of $x^m$ in $$ \left(\sum_{n=1}^{\infty}x^n\right)^3 $$ for some $m\in{\Bbb N}$?

I attempted it using the Cauchy products, but things got messy. I think one might want to write it as $$ \left(\frac{x}{1-x}\right)^3 $$ But I have no idea how to find the Taylor series of this function.

share|cite|improve this question
You should add the "^3" to the question's title – Jean-Sébastien Nov 7 '12 at 15:14
you do realise that the sum is $\left(\cfrac{x}{1-x}\right)^3 \iff |x|<1$ – user31280 Nov 7 '12 at 15:15
up vote 4 down vote accepted

Rewrite this as $x^3(1-x)^{-3}$, write the taylor expansion of $(1-x)^{-3}$, then multiply it by $x^3$

share|cite|improve this answer

Put $x^3(1-x)^{-3} = \sum_{m \ge 0} a_m x^m$, and multiply by $(1-x)^3 = 1-3x+3x^2-x^3$ : you get $x^3 = a_0 + (a_1-3a_0)x^1 + (a_2-3a_1+3_a0)x^2 + \sum_{m \ge 3} (a_m-3a_{m-1}+3a_{m-2}-a_{m-3}) x^m$ , from which you deduce that $a_0 =a_1 = a_2 = 0, a_3 = 1$, and forall $m\ge4$, $a_m = 3a_{m-1}-3a_{m-2}+a_{m-3}$.

This last statement is equivalent to saying that the function $m \mapsto a_m$ is a polynomial in $m$ of degree $2$ on $\{1,2,3,\ldots\}$. Since $a_1 = a_2 = 0$ and $a_3 = 1$ we must have $a_m = \frac {(m-1)(m-2)}{(3-1)(3-2)} = \frac {(m-1)(m-2)}2$ forall $m \ge 1$.

Alternatively, you can just develop the product, observe that $a_m$ is the number of integer triples $(i,j,k)$ such that $i,j,k \ge 1$ and $i+j+k = m$, and try to count them.

share|cite|improve this answer

The usual power series expansion $(1-x)^{-3} = \sum_{k\geq 0} \left( \begin{matrix} -3 \\ k \end{matrix} \right) (-x)^3$ works (in fact for any complex number instead of $-3$). Here, $$ \left( \begin{matrix} -3 \\ k \end{matrix} \right) = \frac{(-3)(-4) \cdots (-3-k+1)}{k!}.$$ Proof by Taylor expansion (as already mentioned).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.