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{

(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)



2Sn=2+3+


两式相减得:
Sn=2+2(



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