Question

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

Answer 1

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)