设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=23,a5+b3=17.(Ⅰ)求{an},{bn}的通项公式;(Ⅱ)设cn=anbn,求数列{cn}的前n项和Sn

题目简介

设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=23,a5+b3=17.(Ⅰ)求{an},{bn}的通项公式;(Ⅱ)设cn=anbn,求数列{cn}的前n项和Sn

题目详情

设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=23,a5+b3=17.
(Ⅰ)求{an},{bn}的通项公式;
(Ⅱ)设cn=anbn,求数列{cn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(I)由已知可得,
1+2d+q4=23
1+4d+q2=17

由等比数列的各项都为正可得,q>0
解可得,q=2,d=3
∴an=1+3(n-1)=3n-2,bn=2n-1
(II)由(1)可得,Cn=(3n-2)•2n-1
Sn=1•20+4•21+…+(3n-2)•2n-1
2Sn=1•2+4•22+…+(3n-5)•2n-1+(3n-2)•2n
两式相减可得,-Sn=1+3(2+22+2n-1)-(3n-2)•2n=1+3×
2(1-2n-1)
1-2
-(3n-2)•2n

Sn=(3n-5)•2n+5

更多内容推荐