Euler's Relation for Planar Graphs

Question

In my class, we're looking at this proof that says "By Euler's relation $3v_{3}+2v_{4}+v_{5}=12+\sum_{k\geq7} (k-6)v_{k}$." I know that there are different ways we can rewrite Euler's formula using summations, but can someone explain how this is true?

Answer 1

Assuming there are no vertices of degree less than $3$, we have $2e = \sum_{k \ge 3} k v_k$ by the handshake lemma.

Euler's formula says that $v - e + f = 2$ I'm assuming that you have a triangulation here, in which case $2e = 3f$ (by the dual of the handshake lemma) and Euler's formula simplifies to $e = 3v-6$. So we have $$ \sum_{k \ge 3} k v_k = 2e = 6v - 12 = \sum_{k\ge 3} 6v_k - 12. $$ This can be rearranged to $$ 12 = \sum_{k \ge 3} (6-k)v_k. $$ That's essentially the expression you've got. The first $4$ terms of the sum on the right are $3v_3$, $2v_2$, $v_5$, and $0v_6$, but after that, everything has a negative coefficient, and it makes sense to move it to the other side.