I am trying to extend the problem from Stanley's book. Has expressed the following problem:
For positive integer $k,n $. Show that: \begin{align} \sum x_1 x_2 \cdots x_k = \binom{n+k-1}{2k-1} \end{align} where the sum ranges over all compositions $(x_1 x_2 \cdots x_k)$ of $n$ into $k$ parts.
I think about extending of this problem as follows:
Problem:
For any positive integers $n$ and integers $a,b$, calculate $$\sum_{x_1+x_2+\cdots+x_k=n}(ax_1+b)\cdots(ax_k+b)$$ I think about this problem with Generating function of $h(x)$ where $$h(x)=\sum_{i\geq 1}(ai+b)x^i$$ Then, we can write $$1+h(x)+h(x)^2+\cdots=\frac{1}{1-h(x)}\text{.}$$ Now, I do not know my method is right or false for answering this qustion!!
Anyway, I have no other idea to solve this question!!
Can someone suggest me an another or better way to solving this question?
In your idea, using of elementary symmetric function for solving this question while the sum range over all partitions of $n$ into at most $k$ parts is useful?
Beforehand, I would like to appreciate your comments and helps!
( NOTE:
I knew this straight result of the elementary symmetric function. A polynomial in $n$ variabels $P(x_1,...,x_n)$ in $C[x_1,...,x_n]$ is known as a symmetric polynomial if any permutation of $S_n$ invariant this polynimial.
An important family of symmetric polynomial is the family of elementary symmetric function $$e_k(x_1,...x_n)=\sum x_{i1}...x_{ik}$$, where $e_1=1$ and zero for $k>n$. It is easy the terms of this polynomial is ${n\choose k}$. The GF of this polynomial is $$E(t)=(1+x_1t)(1+x_2t)...$$, Then If I can write $$(ax_1+b)(ax_2+b)...(ax_k+b)$$. According to $E(t),$ will be very useful for solving this qustion in the case $x_1,x_2,\cdots ,x_k$ are partition of $n$ into $k$ parts!!)
Originally I had misread your question and answered a different question. That answer can be found below. I first answer the question you intended. Let $n$ and $k$ be fixed positive integers. First we compute $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1;}(ax_1+b)\cdots(ax_k+b)\tag{1} $$ where the sum is taken over all compositions of $n$ with $k$ parts. Note that $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}(ax_1+b)\cdots(ax_k+b) =[x^n]\left([(a+b)x+(2a+b)x^2+\dotsb]^k\right)\tag{2}. $$ To simplify the RHS we note that $$ (a+b)x+(2a+b)x^2+(3a+b)x^3+\dotsb=\frac{ax}{(1-x)^2}+\frac{bx}{1-x} =\frac{(a+b)x-bx^2}{(1-x)^2}.\tag{3} $$ Combining (2) and (3) we get that $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}(ax_1+b)\cdots(ax_k+b) =[x^n]\left(\frac{[(a+b)x-bx^2]^{k}}{(1-x)^{2k}}\right) \tag{4} $$ which we can compute. As a sanity check consider the special case where $a=1$ and $b=0$. Then $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}x_1\cdots x_k =[x^n]\left(\frac{x^{k}}{(1-x)^{2k}}\right) =[x^{n-k}]\left(\frac{1}{(1-x)^{2k}}\right) =\binom{n+k-1}{2k-1} $$ by the extended binomial theorem. This answer agrees with Stanley's answer. Returing to 4 and applying the binomial theorem in the numerator we get that $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}x_1\cdots x_k =\sum_{m=0}^{k} \binom{k}{m}\binom{n+k-m-1}{2k-1}(-1)^m (a+b)^{k-m}b^m \tag{5} $$ Here is my original answer. In the following we fix $n$ and compute $$\sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1;\, k\geq 1}(ax_1+b)\cdots(ax_k+b)$$ where the sum is taken over all compositions of $n$. We suppress $k \geq 1$ in the subscript of the sums in what follows. Note that $$\sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}(ax_1+b)\cdots(ax_k+b) =[x^n]\left(\sum_{k\geq 1} [(a+b)x+(2a+b)x^2+\dotsb]^k\right)\tag{1} $$ where $[x^n]$ extracts the coefficient of $x^n$ in the formal power series. To simplify the right hand side we notice that $$ (a+b)x+(2a+b)x^2+(3a+b)x^3+\dotsb=\frac{ax}{(1-x)^2}+\frac{bx}{1-x} =\frac{(a+b)x-bx^2}{(1-x)^2}.\tag{2} $$ Combining (1) and (2) we get that $$ \sum_{k\geq 1} [(a+b)x+(2a+b)x^2+\dotsb]^k =\frac{\frac{(a+b)x-bx^2}{(1-x)^2}}{1-\frac{(a+b)x-bx^2}{(1-x)^2}} =\frac{(a+b)x-bx^2}{1-(2+a+b)x+(b+1)x^2}.\tag{3} $$ Hence it follows from (1) that $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}(ax_1+b)\cdots(ax_k+b) =[x^n]\left(\frac{(a+b)x-bx^2}{1-(2+a+b)x+(b+1)x^2}\right) $$ which you can compute.
As a special case consider $a=1$ and $b=0$ so that $$ \sum_{x_1+x_2+\cdots+x_k=n;\,x_i\geq 1}x_1\cdots x_k =[x^n]\left(\frac{x}{1-3x+x^2}\right) =F_{2n} $$ which agrees with Stanley's solution.