问题 解答题

(14分)已知数列{an}的前n项和为Sn,且an是Sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上。  (1)求a1和a2的值;  (2)求数列{an},{bn}的通项an和bn

答案

(1)2,4(2)bn=2n-1,an=2

(1)∵anSn与2的等差中项∴Sn=2an-2              。。。。1

a1=S1=2a1-2,解得a1="2              " 。。。。2

a1+a2=S2=2a2-2,解得a2="4     " 。。。        。3

(2)∵Sn=2an-2,Sn-1=2an-1-2,

SnSn-1=an                。。。。5

an=2an-2an-1,    ∵an≠0,∴,。。8

即数列{an}是等比数列∵a1=2,∴an=2n                                             10

∵点P(bnbn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,。。。       12

bn+1-bn=2,即数列{bn}是等差数列,                      13

b1=1,∴bn=2n-1,                                   14分

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