you've only shown 3 out of the many, many cases for k odd, and 4 cases for k even. this particular proof is a pain, because the number of cases grows rapidly as k does.
let's say S is a sum (associated arbitrarily) of k+1 vectors. we want to show that S = ((...(v1+v2)+v3)+...+vk)+vk+1. we will assume this is true for any sum of j vectors with: 3 ≤ j ≤ k.
we can write:
S = A+B, for two sums A and B, and the largest either A or B can be is a sum of k vectors.
applying our induction hypothesis to B we have:
B = ((...(vj+vj+1)+vj+2)+...+vk)+vk+1, for some vj.
applying our induction hypothesis to A, we have:
A = ((...(v1+v2)+v3)+...+vj-2)+vj-1.
if we let C = ((...(vj+vj+1)+...+vk-2)+vk,
we have A+B = A+(C+vk+1) = (A+C)+vk+1.
applying the induction hypothesis to A+C, we have:
A+C = ((...(v1+v2)+v3+...+vk-1)+vk,
hence S = (A+C)+vk+1 = ((...(v1+v2)+v3)+...+vk)+vk+1, as desired.
(the problem with your proof is that association can occur with more than "adjacent pairs", you can have sums like:
(v1+(v2+v3)+v4)+v5 and so on).