已知数列｛an}是公差不为零的等差数列且a1=1,且a2,a4,a8成等比数列.①求通项an

题目

已知数列｛an}是公差不为零的等差数列且a1=1,且a2,a4,a8成等比数列.①求通项an
②若bn=1/[an*a(n+1)],求｛bn｝的前n项和Sn

答案

设an=a1+(n-1)d
即an=1+(n-1)d
a2,a4,a8成等比数列
(1+3d)^2=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d
d≠0 d=1
an=1+(n-1)=n
bn=1/[n(n+1)]
Sn=1/(1*2)+1/(2*3)+1/(3*4)+.+ 1/[n(n+1)]
=1/1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)