Question

已知数列{an}是一个等差数列,且a2=5,a5=11,令bn=1/(a^2n-1),求数列{bn}的前n项和

Answer 1

数列{an}是一个等差数列,且a2=5,a5=11
所以，3d＝a5－a2＝11－5＝6
d＝2
a1＝a2－d＝5－2＝3
所以，
an＝3+2(n-1)＝2n+1

bn=1/(an²-1)
=1/[(2n+1)²-1]
=1/[4n(n+1)]
=(1/4)×[1/n－1/(n+1)]

数列{bn}的前n项和
=(1/4)×[1-1/2+1/2-1/3+……1/n-1/(n+1)]
=(1/4)×[1－1/(n+1)]
=n/[4(n+1)]

Answer 2

求出an=2n+1,则bn=1/[(2n+1)^2n-1]=1/2{1/[(2n+1)^n-1]-1/[(2n+1)^n+1]},相加相消得
Sn=1/2{1/2-1/[(2n+1)^n+1]

Answer 3

[(4n+1)n-1/(4n+1)n]^n /[(4n+1)n-1]