Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is my question:

Given $n$ and $k$, find the coefficient of $x^n$ in $(1+x+x^2+x^3+\cdots)^k$

The largest term needed in each block is $n-k+1$, and that's all I have figured out. I'm guessing the answer includes $n-k+1$ choose something.

I don't even know if I can solve this, but any help would be appreciated!

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

One failproof way of doing this using the generating function:

$$\frac{1}{1-x}=1+x+x^2+\ldots$$

So you are interested in the $n$'th coefficient of $1/(1-x)^k$ which amounts to taking $n$ derivatives and evaluating at $x=0$, then multiplying by 1/n!:

$$\frac{1}{n!}\left.\frac{d^n }{dx^n}(1-x)^{-k}\right|_{x=0}=\frac{1}{n!}k(k+1)\cdots (k+n-1)=\frac{(k+n-1)!}{n!(k-1)!}$$

Gee, that looks awfully like $\binom{k+n-1}{n}$ (or $\binom{k+n-1}{k-1}$ depending on your tastes)... Perhaps that will give you a hint of a direct combinatorics approach.

share|improve this answer
add comment

We have

$(1+x+x^2+x^3+\cdots)^k = (1+x+x^2+ \cdots) (1+x+x^2 + \cdots)\cdots(1+x+x^2+ \cdots)$

where there are $k$ of the $(1 + x + x^2 + \cdots)$.

To form the term $x^n$, we need to pick $\displaystyle x^{j_i}, 1 \leq i \leq k$ from each of the $k$ factors, such that $j_1 + j_2 + j_3 + \cdots + j_k = n$. This is a partition of $n$ into $k$ parts. The number of ways to do this is known to be $\displaystyle {n+k-1 \choose k-1}$.

share|improve this answer
add comment

Your Answer

 
discard

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.