If $a$ is in natural number, show that the number of positive integral solutions of $$x_1+2x_2+3x_3+\dots+nx_n= a$$ is equal to the number of nonnegative integral solution of $$y_1+2y_2+3y_3+\dots+ny_n=a-\frac{n(n+1)}{2}.$$
I tried to find the number of solutions of both by using the formula $\binom{n+r-1}{r-1}$ and I equated them after this I am not able to proceed. Please tell if my approach is correct if not what is the correct solution.
Any help will be appreciated , thanks in advance.
There is a bijection between the two sets of solutions. $(x_1,x_2,\dots,x_n)$ is a positive integer solution of the first equation iff $(y_1,y_2,\dots,y_n)$ with $y_i=x_i-1$ for $i=1,\dots,n$ is a nonnegative integer solution of the second equation.
The number of solutions is the coefficient of $x^a$ in \begin{eqnarray*} \frac{x^{\frac{n(n+1)}{2}}}{(1-x)(1-x^2) \cdots (1-x^n) }. \end{eqnarray*}
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The first OP statement is given by the enclosed expression at the very beginning:
\begin{align} &\bbx{\sum_{x_{1} = \color{#f00}{\large 1}}^{\infty} \sum_{x_{2} = \color{#f00}{\large 1}}^{\infty} \sum_{x_{3} = \color{#f00}{\large 1}}^{\infty}\cdots \sum_{x_{n} = \color{#f00}{\large 1}}^{\infty} \bracks{z^{a}}z^{\large \,x_{1} + 2x_{2} + 3x_{3} + \cdots + nx_{n}}} \\[5mm] = &\ \sum_{x_{1} = \color{#f00}{\large 0}}^{\infty} \sum_{x_{2} = \color{#f00}{\large 0}}^{\infty} \sum_{x_{3} = \color{#f00}{\large 0}}^{\infty}\cdots \sum_{x_{n} = \color{#f00}{\large 0}}^{\infty}\bracks{z^{a}} z^{\large \,\pars{x_{1} + 1} + 2\pars{x_{2} + 1} + 3\pars{x_{3} + 1} + \cdots + n\pars{x_{n} + 1}} \\[5mm] = &\ \sum_{x_{1} = \color{#f00}{\large 0}}^{\infty} \sum_{x_{2} = \color{#f00}{\large 0}}^{\infty} \sum_{x_{3} = \color{#f00}{\large 0}}^{\infty}\cdots \sum_{x_{n} = \color{#f00}{\large 0}}^{\infty} \bracks{z^{a}}z^{\large n\pars{n + 1}/2\ +\ \,x_{1} + 2x_{2} + 3x_{3} + \cdots + nx_{n}} \\[5mm] = &\ \bbx{\sum_{y_{1} = \color{#f00}{\large 0}}^{\infty} \sum_{y_{2} = \color{#f00}{\large 0}}^{\infty} \sum_{y_{3} = \color{#f00}{\large 0}}^{\infty}\cdots \sum_{y_{n} = \color{#f00}{\large 0}}^{\infty}\bracks{z^{a - n\pars{n + 1}/2}} z^{\large \,y_{1} + 2y_{2} + 3y_{3} + \cdots + ny_{n}}} \end{align}
which is the OP second statement.
