已知等比数列{an}的前n项和为Sn,且an是Sn与2的等差中项,等差数列{bn}中,b1=2,点P(bn,bn+1)在直线y=x+2上求a1和a2的值求数列{an}和{bn}的通项an和bn设Cn=an*bn,求数列{cn}的前n项和Tn
问题描述:
已知等比数列{an}的前n项和为Sn,且an是Sn与2的等差中项,等差数列{bn}中,b1=2,点P(bn,bn+1)在直线y=x+2上
求a1和a2的值
求数列{an}和{bn}的通项an和bn
设Cn=an*bn,求数列{cn}的前n项和Tn
答
(2)∵Sn=2an-2,Sn-1=2an-1-2,
又Sn-Sn-1=an,n≥2
∴an=2an-2an-1,
∵an≠0,
∴anan-1=2(n≥2),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=2,∴bn=2n
bn应该这样
答
(1)由an是Sn与2的等差中项得2an=Sn +2n=1时,2a1=S1+2=a1+2 a1=2n=2时,2a2=S2+2=a1+a2+2 a2=a1+2=2+2=4(2)n=1时,a1=2n≥2时,2an=Sn+2 2a(n-1)=S(n-1)+22an-2a(n-1)=Sn+2 -S(n-1)-2=Sn-S(n-1)=anan=2a(n-1)an/a(n-1)=2...