Suppose $v_1,v_1+v_2,v_1+v_2+v_3,...,\sum_{i=1}^nv_i$ are linear independent. Show that $v_1,v_2,...,v_n$ are also linear independent. 
Suppose $v_1,v_1+v_2,v_1+v_2+v_3,...,\sum_{i=1}^nv_i$ are linear independent. Show that $v_1,v_2,...,v_n$ are also linear independent. 

Using the definition of linear independence, $$\alpha_1v_1+\alpha_2(v_1+v_2)+\alpha_3(v_1+v_2+v_3)+...+\alpha_n\sum_{i=1}^nv_i=0,$$ only if $\alpha_1=\alpha_2=...=\alpha_n=0$.
So $\sum_{i=1}^n\alpha_i\sum_{j=1}^iv_j=0$, if I'm correct. 
Therefore $\sum_{i=1}^n\sum_{j=1}^i\alpha_iv_j=0$
I got stuck here and I don't know how to continue.

By induction on $n.$ 
(i). $n=1.$ Trivial.
(ii). If true for  case $n,$ let $u_1=v_1, u_2=v_1+v_2,..., u_{n+1}=v_1+...+v_{n+1}$ where $u_1,...,u_{n+1}$ are $n+1$ linearly independent vectors.. Note that $v_j=u_j-u_{j-1}$ for $1<j\leq n+1.$ So $$\sum_{i=1}^{n+1}K_iv_i=0\implies K_1u_1+\sum_{j=2}^{n+1} K_j(u_j-u_{j-1})=0.$$ The co-efficient of $u_{n+1}$ in the above sum is $K_{n+1}.$ By the independence of $u_1,...,u_{n+1}$ we have $K_{n+1}=0.$ Therefore $$\sum_{i=1}^{n+1}K_iv_i=\sum_{i=1}^nK_iv_i.$$ Since $u_1,...,u_n$ are $n$ linearly independent vectors and the result holds for case $n,$ we have $K_i=0$ for $1\leq i\leq n.$
