Prove that the sum over triples of non-negative integers which sum to n of $(-1)^{n_1}$ equals one. Problem Statement:
Prove that for any positive integer $n$, $$\sum_{n_1+n_2+n_3 = n} (-1)^{n_1} = 1$$ where the summation is over all triples $(n1, n2, n3)$ of non-negative integers with sum $n$.
What I did so far:
I tried applying Multinomial Theorem by letting $x_1 = -1, x_2=x_3=1$, and ended up with:
$$(-1+1+1)^n = \sum_{\substack{n_1+n_2+n_3 = n \\ n_1,n_2,n_3 \geq 0}} {n \choose{n_1, n_2, n_3} }(-1)^{n_1} $$ but I am unsure how to deal with the multinomial coefficient. Alternatively, this sum can be restated in terms of Binomial Theorem, because $n_2+n_3 = n-1$:
$$(-1+2)^n = \sum_{n_1=0}^n {n \choose {n_1}}(-1)^{n_1}\cdot(2)^{n-n1} = 1$$ but this doesn't seem to help. Any tips for how to approach this?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Prove that for $\ds{\,\underline{any\ positive\ integer}\,}$ $\ds{}$,
$\ds{\bbox[5px,#ffd]{%
\sum_{n_{1} + n_{2} + n_{3} = n}\pars{-1}^{n_{1}} = 1}:\
{\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{%
\sum_{\substack{%
n_{1} + n_{2} + n_{3}\ =\ n
\\[0.5mm]
n_{1}, n_{2}, n_{3}\ \in\
\mathbb{N}_{\color{red}{\geq 1}}}}\pars{-1}^{n_{1}}} =
\sum_{n_{1} = 1}^{\infty}\sum_{n_{2} = 1}^{\infty}
\sum_{n_{3} = 1}^{\infty}\pars{-1}^{n_{1}}
\bracks{z^{n}}z^{n_{1}\ +\ n_{2}\ +\ n_{3}}
\\[5mm] = &\
\bracks{z^{n}}\sum_{n_{1} = 1}^{\infty}\pars{-z}^{n_{1}}
\sum_{n_{2} = 1}^{\infty}z^{n_{2}}
\sum_{n_{3} = 1}^{\infty}z^{n_{3}} =
\bracks{z^{n}}{z \over 1 - \pars{-z}}\,{z \over 1 - z}\,
{z \over 1 - z}
\\[5mm] = &\
\bracks{z^{n - 3}}\pars{1 - z^{2}}^{-1}\pars{1 - z}^{-1} =
\bracks{z^{n - 3}}\sum_{i = 0}^{\infty}z^{2i}
\sum_{j = 0}^{\infty}z^{j}
\\[5mm] = &\
\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\bracks{j = n - 3 - 2i} =
\sum_{i = 0}^{\infty}\bracks{n - 3 - 2i \geq 0} =
\sum_{i = 0}^{\infty}
\bracks{i \leq \left\lfloor\,{n - 3 \over 2}\,\right\rfloor}
\\[5mm] = &\
\bbx{\left\lfloor\,{n - 3 \over 2}\,\right\rfloor + 1} \\ &
\end{align}

A: As it turns out, there was a typo in this question. The correct prompt would be:
Prove that for any positive integer $n$, $$\sum_{n_1+n_2+n_3 = n} (-1)^{n_1} {n \choose {n_1,n_2,n_3}} = 1$$ where the summation is over all triples $(n_1, n_2, n_3)$ of non-negative integers with sum $n$.
This I know how to prove: Use the definition of Multinomial Theorem to represent the quantity $(x_1+x_2+x_3)^n$. Let $x_1=-1, x_2=x_3=1$ and the result should be the above sum.
Thank you all for your answers / comments! I appreciate your replies, even though the question turned out to be ill-formed
