等比数列中怎样确定项数?

Question

如题所述

Answer 1

用末项除以首项，得到商记为m，q为公比，项数n=logq(m),q为底数

Answer 2

an=a1*q^(n-1)
q^(n-1)=an/a1
所以n=1+logq(an/a1)本回答被提问者采纳

Answer 3

原式= a²+ab+2ab+2b²
=a(a+b)+2b(a+b)
=(a+b)(a+2b)=[2^(n+4)-2^(n+1)]÷[2^(n+4)]
=2^(n+4)÷[2^(n+4)]-2^(n+1)÷[2^(n+4)]
=1-2^(n+1-n-4)
=1-2^(-3)
=1-1/2³
=7/8

Answer 4