Let i1,i2,…,ip be vertices of X such that {i1,i2},{i2,i3},…,{ip−1,ip} are edges of order 3. Then q(∑r=1preir)=21p(p+1).
We have (eir∣eir)=1,(eir∣eir+1)=−21,(eir∣eis)=0 if s>r+1. Thus
q(r=1∑preir)=r=1∑pr2−2r=1∑p−121r(r+1)=p2−r=1∑p−1r.
By Theory of Sets, Chap. III. § 5, no. 8. Cor. to Prop. 14, this is equal to
p2−21p(p−1)=21p(p+1).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.