Sum of products of solutions of a diophantine equation.

Question

I have problems solving the following:

There is an equation $y_1+y_2+...+y_k=n$

Show that $\displaystyle \sum_{y_1,...,y_k} y_1 \cdot y_2 \cdot ... \cdot y_k $

equals the coefficient of $x^n$ in $(x+2x^2+3x^3+...)^k$

This comes from Victor Bryant's "Aspects of combinatorics"

I know that we can find the number of solutions of the equation above by looking for the number of ways we can get $x^n$ in $(1+x+x^2+...)(1+x+x^2+...)...(1+x+x^2+...)$ (there are $k$ brackets).

Could you help me?

Thanks.

Answer 1

Hint: Since you already know the bijection which states that the number of solutions are the same, think about the exact pairing that is given, but use the expansion of

$$ (1 + x_1 + x_ 2^2 + \ldots ) (1 + x_2 + x_2 ^2 + \ldots ) (1 + x_k + x_k^2 + \ldots) $$

Now, what happens to each term of degree $n$ when we differentiate this expression with respect to each of the terms? I.e. $\frac {d}{dx_1} \frac d{dx_2} \ldots \frac {d}{dx_k}$. Relate that to the corresponding $\prod y_i$.