設{an}公差為d,{bn}公差為d'
lim(an/bn)
=lim[(a1+(n-1)d]/[b1+(n-1)d']
=lim[(a1-d)+nd]/[(b1-d')+nd']
=lim[(a1-d)/n +d]/[(b1-d')/n +d']
a1-d,b1-d'均為定值,n->+∞,(a1-d)/n->0 (b1-d')/n->0
lim(an/bn)=d/d',又lim(an/bn)=3,因此d/d'=3 以上即為得到公差比的詳細步驟.
lim[(b1+b2+...+b3n)]/[n×a(2n)]
=lim[3nb1+3n(3n-1)d'/2]/[n×a(2n)]
=lim[3b1+3(3n-1)d'/2]/[a1+(2n-1)d]
=lim[6b1+3(3n-1)d']/[2a1+(4n-2)d]
=lim[(6b1-3d')+9nd']/[(2a1-2d)+4nd]
=lim[(6b1-3d')/n +9d']/[(2a1-2d)/n +4d]
=9d'/4d
=(9/4)[1/(d/d')]
=(9/4)(1/3)
=3/4