Question

求解高中数列题

等差数列｛An}的各项均为正数，a1=3，前n项和为Sn，｛Bn}为等比数列,b1=1，且b2s2=64，b3s3=960.（1）求通项An，Bn。（2）求和1/s1+1/s2+1/s3+……+1/sn怎么做呀

Answer 1

(1)
设{an}的公差为d,{bn}的公比为q
∴an=3+(n-1)d,bn=q^(n-1)
∵b2s2=64
∴q(a1+a2)=64
==>(6+d)*q=64 ①
∵ b3s3=960
∴q²(a1+a2+a3)=960
==> (9+3d)*q²=960
==> (3+d)*q²=320 ②
①²/②:
(6+d)²/(3+d)=64²/320=64/5
==>5( 36+12d+d²)=192+64d
==>5d²-4d-12=0
==>d=2或d=-6/5(舍去）
∴q=8
∴an=2n+1,bn=8^(n-1)

(2)
Sn=(3+2n+1)*n/2=n(n+2)
∴1/Sn=1/[n(n+2)]
=1/2[1/n-1/(n+2)]
∴1/s1+1/s2+1/s3+……+1/sn
=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+........+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=3/4-(2n+3)/[2(n+1)(n+2)]

Answer 2

(1)根据题意：设an=a1+（n-1）d，bn=b1*q^（n-1）
可得方程组
(3+3+d)*q=64
(3+3+d+3+2d)*q^2=960
d=2,q=8
∴an=3+（n-1）*2=2n+1
bn=8^（n-1）
(2)sn= n^2+2n
1/s1+1/s2+1/s3+……+1/sn和为S
S=1/3+1/8+1/15+1/24…+1/(n*(n+2))
=1/2（1-1/3+1/2-1/4+1/3-1/5… +1/n-1/(n+2))
=1/2（3/2-1/(n+1）-1/（n+2））
=3/4-(2n+3）/(2*(n+1）*（n+2））