Linear combination of of $n+2$ vectors Given $n+2$ vectors in an $n$-dimensional space, prove that there exists a non-trivial linear combination $$a_1 v_1 + a_2v_2 + \cdots + a_{n+2}v_{n+2}=0$$ with not all $a_i = 0$, yet with $\sum\limits_{i=1}^{n+2} a_i = 0$.

I tried to take $n+1$ vectors and build a barycentric combination of them, such that they would form the $(n+2)$nd vector and $\sum\limits_{i=1}^{n+1} a_i = 1$, and if I add $(n+2)$nd vector to this combination with $a_{n+2}=-1$, it will give the needed linear combination.
But I stuck proving this barycentric combination exists. So now I think I'm moving in the wrong direction, because it seems, that there is can be no barycentric combinations formed from these vectors.
 A: The set $\{v_1, \ldots, v_{n+2}\}$ is linearly dependent so there is a vector which can be expressed as a linear combination of the remaining $n+1$ vectors. WLOG assume
$$v_{n+2}=\sum_{i=1}^{n+1} \alpha_i v_i.$$
On the other hand, the set $\{v_1, \ldots, v_{n+1}\}$ is also linearly dependent so there are scalars $\beta_1, \ldots, \beta_{n+1}$ not all zero such that
$$\sum_{i=1}^{n+1} \beta_iv_i = 0.$$
Notice that if $\sum_{i=1}^{n+1} \beta_i = 0$ we are done as
$$\sum_{i=1}^{n+1} \beta_iv_i +0\cdot v_{n+2}= 0$$
would be our desired linear combination. Therefore we can assume $\sum_{i=1}^{n+1} \beta_i \ne 0$ and we have for all $\gamma \in \Bbb{R}$ that
$$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i )v_i - v_{n+2}=\underbrace{\sum_{i=1}^{n+1} \alpha_i v_i - v_{n+2}}_{=0}  + \gamma\underbrace{\sum_{i=1}^{n+1} \beta_iv_i}_{=0} = 0$$
where the sum of these scalars is
$$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i ) - 1 = \sum_{i=1}^{n+1}\alpha_i + \gamma \sum_{i=1}^{n+1} \beta_i - 1$$
so picking $\gamma = \frac{1-\sum_{i=1}^{n+1}\alpha_i}{\sum_{i=1}^{n+1} \beta_i}$ gives the desired linear combination.
