题目内容
(本小题满分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;
(1)求a1和a2的值; (2)求数列{an},{bn}的通项an和bn;
解:(1)∵an是Sn与2的等差中项 ∴Sn=2an-2 ∴a1=S1=2a1-2,解得a1=2
a1+a2=S2=2a2-2,解得a2="4"
(2)∵Sn=2an-2,Sn-1=2an-1-2,又Sn—Sn-1=an, ∴an=2an-2an-1, ∵an≠0,
∴,即数列{an}是等比数列 ∵a1=2,∴an=2n
∵点P(bn,bn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,
∴bn+1-bn=2,即数列{bn}是等差数列,又b1=1,∴bn=2n-1
a1+a2=S2=2a2-2,解得a2="4"
(2)∵Sn=2an-2,Sn-1=2an-1-2,又Sn—Sn-1=an, ∴an=2an-2an-1, ∵an≠0,
∴,即数列{an}是等比数列 ∵a1=2,∴an=2n
∵点P(bn,bn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,
∴bn+1-bn=2,即数列{bn}是等差数列,又b1=1,∴bn=2n-1
略
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