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=21,a5+b3=13 (Ⅰ)求{an},{bn}的通项公式; (Ⅱ)求数列{an•bn}的前n项和Sn.
问题描述:
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=21,a5+b3=13
(Ⅰ)求{an},{bn}的通项公式;
(Ⅱ)求数列{an•bn}的前n项和Sn.
答
(I)设{an}的公差为d,{bn}的公比为q,则依题意有q>0,
∵a1=b1=1,a3+b5=21,a5+b3=13,
∴
,解得d=2,q=2.
1+2d+q4=21 1+4d+q2=13
∴an=1+(n-1)d=2n-1,bn=2n−1,
(Ⅱ)由(I)得,an•bn=(2n-1)•2n-1,
Sn=1•20+3•21+…+(2n-1)•2n-1
2Sn=1•2+3•22+…+(2n-3)•2n-1+(2n-1)•2n
两式相减可得,-Sn=1+2(2+22+2n-1)-(2n-1)•2n
=1+2×
-(2n-1)•2n2(1−2n−1) 1−2
=(3-2n)•2n-3,
则Sn=(2n-3)•2n+3.