The set $(v_1,v_2,...,v_n)$ is a basis for the $\mathbb R$-vector space $V$. Now $u_1=v_1$, $u_2=v_1+v_2$, $u_n=v_1+v_2+,...,+v_n$. Show that the set $U:=(u_1,u_2,..u_n)$ is also a basis of $V$.
Proof : $U$ is a basis of $V$, if it's both linearly independent and such that it spans $V$. To show linear independence, it should be sufficient to look at the matrix with columns $$(u_1,u_2,u_3,...,u_n)$$
Which is obviously in row echelon form and there are no free variables. As such the homogeneous equation $Ux=0$ has only the trivial solution and $U$ is linearly independent. Showing that $U$ spans $V$ is the problem. I believe that, since the dimension of $V$ is $n$ and is isomorphic to $\Bbb{R^n}$, every set that spans $V$ must have exactly $n$ vectors, which are linearly independent. This is the case with $U$ - what else must be shown in order to verify that $U$ indeed spans $V$?