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 = ((...(v_{1}+v_{2})+v_{3})+...+v_{k})+v_{k+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 = ((...(v_{j}+v_{j+1})+v_{j+2})+...+v_{k})+v_{k+1}, for some v_{j}.

applying our induction hypothesis to A, we have:

A = ((...(v_{1}+v_{2})+v_{3})+...+v_{j-2})+v_{j-1}.

if we let C = ((...(v_{j}+v_{j+1})+...+v_{k-2})+v_{k},

we have A+B = A+(C+v_{k+1}) = (A+C)+v_{k+1}.

applying the induction hypothesis to A+C, we have:

A+C = ((...(v_{1}+v_{2})+v_{3}+...+v_{k-1})+v_{k},

hence S = (A+C)+v_{k+1}= ((...(v_{1}+v_{2})+v_{3})+...+v_{k})+v_{k+1}, as desired.

(the problem with your proof is that association can occur with more than "adjacent pairs", you can have sums like:

((v_{1}+v_{2})+v_{3})+(v_{4}+v_{5}) or

(v_{1}+(v_{2}+v_{3})+v_{4})+v_{5}and so on).