Question

已知数列{an},{bn}分别是等差，等比数列，且a1=b1=1,a2=b2,a4=b3≠b4

（1）求数列{an}, {bn}的通项公式；
（2）设Sn为数列{an}的前n项和，求数列{ 1/Sn }的前n项和R

Answer 1

a2=b2
a1+d=b1q
d+1=q

a1+3d=b1q^2
1+3d=q^2
(d+1)^2=3d+1
d^2+2d+1=3d+1
d^2-d=0
d(d-1)=0
d=0或d=1
当d=1时
q=1+1=2
当d=0时
q=1(舍去，b3≠b4)
所以d=1,q=2
an=a1+(n-1)d
=1+n-1
=n

bn=b1q^(n-1)
=1*2^(n-1)
=2^(n-1)

sn=1+2+3+.......+n=n(n+1)/2
1/sn=1/[n(n+1)/2]
=2/[n(n+1)]
=2*[1/n-1/(n+1)]

R=2*(1-1/2)+2*(1/2-1/3)+..........+2*[1/n-1/(n+1)]
=2*[1-1/2+1/2-1/3+....+1/n-1/(n+1)]
=2*[1-1/(n+1)]
=2n/(n+1)