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.
人气:342 ℃ 时间:2019-08-19 08:28:39
解答
(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,∴1+2d+q4=211+4d+q2=13,解得d=2,q=2. ∴an=1+(n-1)d=2n-1,bn=2n−1,(Ⅱ)由(I)得,an•bn=(2n-1)•2n-1...
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