Is this Set of Vectors Linearly Independent in $V$? Decide whether the given set of vectors is linearly independent in the indicated vector space:
$\{ x_1, x_1 +x_2, x_1 +x_2 +x_3, ..., x_1+\cdots+x_n\} $
if $\{x_1, x_2, x_3, ..., x_n\}$ is linearly
independent, in some vector space $V$.

If $n=4:$
$x_1 - (x_1+x_2) + (x_1+x_2+x_3) - (x_1+x_2+x_3+x_4) = -x_4.$
So, if $n$ is even then it's linearly independent right?

If $n=3:$ 
$x_1 - (x_1+x_2) + (x_1+x_2+x_3) = x_1 + x_3.$
What about this situation when $n$ is odd? What can we state from $x_1+x_3$?
 A: It's best that you see this for yourself, so I'll give just a few hints:
$(1)$ Let $v_1,\ldots, v_n\in V$. If $\dim \text{span}(v_1,\ldots, v_n)=n$, what can we say about the linear independence (or dependence) of the set $\{v_1,\ldots, v_n\}$?
$(2)$ If we are given another family of vectors $\{x_1,\ldots, x_n\}$ and we have that $x_j\in \text{span}(v_1,\ldots, v_n)$ for each $1\le j\le n$, what does this tell us about the relationship between $\text{span}(x_1,\ldots, x_n)$ and $\text{span}(v_1,\ldots, v_n)$? 
$(3)$ If we have subspaces $U\subseteq W\subseteq V$, then what can we say about $\dim(U)$ and $\dim(W)$?
$(4)$ Returning to the original question: Observe that 
$$ (x_1+\cdots +x_k)-(x_1+\cdots+x_{k-1})=x_k$$
for all $2\le k\le n$.
Can you take it from here?
A: Let suppose that $V$ is a vector space over a field $F$. Consider the linear combination
$$\lambda_1\cdot x_1+\lambda_2\cdot (x_1+x_2)+\cdots+\lambda_n\cdot(x_1+x_2+\cdots+x_n)=0$$
where $\lambda_1,\ldots,\lambda_n\in F$
so,
$$(\lambda_1+\lambda_2+\cdots+\lambda_n)\cdot x_1+(\lambda_2+\cdots+\lambda_n)\cdot x_2+\cdots+(\lambda_{n-1}+\lambda_n)\cdot x_{n-1}+\lambda_n\cdot x_n=0$$
The fact that $\{x_1,x_2,\ldots,x_n\}$ are linearly independent gives you the following linear system:
\begin{eqnarray}
\lambda_1+\lambda_2+\cdots+\lambda_{n-1}+\lambda_n &=0 \\
\lambda_2+\cdots+\lambda_{n-1}+\lambda_n&=0 \\
\vdots \qquad \vdots & \\
\lambda_{n-1}+\lambda_n&=0\\
\lambda_n&=0
\end{eqnarray}
No you can to show that the above system has unique solution. Using induction on $n$ for instance. You can also consider the associated matrix of the system and see that it's determinant is always 1, so your set $\{x_1,x_1+x_2,\ldots,x_1+\cdots+x_n\}$ is always linearly independent provided that $\{x_1,\ldots,x_n\}$ is linearly independent.
