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 $x_1+x_3$?
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.