题目内容
(本小题满分12分)
设{an}是等差数列,{bn}是各项为正项的等比数列,且a1=b1="1," a3+b5="21," a5+b3=13.
(1)求{an}, {bn}的通项公式;
(2)求数列{}的前n项和Sn;
设{an}是等差数列,{bn}是各项为正项的等比数列,且a1=b1="1," a3+b5="21," a5+b3=13.
(1)求{an}, {bn}的通项公式;
(2)求数列{}的前n项和Sn;
(1)an="2n-1, " bn=2n-1(2)
解:(1)设{an}的公差为d,{bn}的公比为q,则依题意有q>0,
解得d="2,q=2." 所以an="2n-1, " bn=2n-1
((2), Sn=1+
2Sn=2+3+
两式相减得:
Sn=2+2(=2+
解得d="2,q=2." 所以an="2n-1, " bn=2n-1
((2), Sn=1+
2Sn=2+3+
两式相减得:
Sn=2+2(=2+
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