Question

已知等差数列{An}的各项均为正数,a1=3,前n项和为Sn,{Bn}是等比数列,b1=1,b2S2=64,b3S3=9

Answer 1

(1)

an=a1+(n-1)d

bn=b1.q^(n-1)

a1=3,b1=1

Sn = a1+a2+...+an

b2.S2=64

q(6+d)=64 (1)

b3.S3=960

q^2.(3+d)3=960

q^2.(3+d)=320 (2)

(2)/(1)^2

(3+d)/(6+d)^2= 5/64

64(3+d)= 5(6+d)^2

5d^2-4d-12 =0

(d-2)(5d+6)=0

d=2

from (1) ,q=8

an = 3+2(n-1)=2n+1

bn=8^(n-1)

(2)

Sn = a1+a2+...+an

= (n+2)n

1/Sn = (1/2)[1/n - 1/(n+2)]

1/S1+1/S2+...+1/Sn

=(1/2)[ 1+ 1/2 - 1/(n+1) -1/(n+2) ]

< (1/2)(1+1/2)

= 3/4