Number of solutions being equal 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.
 A: 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.
A: 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*}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.

