记数列{an}的前n项和Sn,且Sn=c/2*n^2+(1-c/2)n(c为常数,n属于N*),且a1,a2,a5成公比
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记数列{an}的前n项和Sn,且Sn=c/2*n^2+(1-c/2)n(c为常数,n属于N*),且a1,a2,a5成公比不等于1的等比
数列
(1)求c的值
(2)设bn=1/anan+1,求数列{bn}的前n项和Tn
数列
(1)求c的值
(2)设bn=1/anan+1,求数列{bn}的前n项和Tn
![记数列{an}的前n项和Sn,且Sn=c/2*n^2+(1-c/2)n(c为常数,n属于N*),且a1,a2,a5成公比](/uploads/image/z/18891449-17-9.jpg?t=%E8%AE%B0%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%2C%E4%B8%94Sn%3Dc%2F2%2An%5E2%2B%281-c%2F2%29n%EF%BC%88c%E4%B8%BA%E5%B8%B8%E6%95%B0%2Cn%E5%B1%9E%E4%BA%8EN%2A%EF%BC%89%2C%E4%B8%94a1%2Ca2%2Ca5%E6%88%90%E5%85%AC%E6%AF%94)
(1)
n=1时,a1=S1=(c/2)×1²+(1- c/2)×1=1
n≥2时,an=Sn-S(n-1)=(c/2)×n²+(1- c/2)n-[(c/2)×(n-1)²+(1- c/2)×(n-1)]=nc-c+1
n=1时,a1=c-c+1=1,同样满足通项公式
数列{an}的通项公式为an=nc-c+1
a1、a2、a5成等比,则a2²=a1·a5
(2c-c+1)²=1·(5c-c+1)
整理,得c²-2c=0
c(c-2)=0
c=0或c=2
c=0时,an=1 a1=a2=a5,公比为1,与已知矛盾,舍去
c=2
(2)
c=2代入{an}通项公式
an=2n-2+1=2n-1
bn=1/[ana(n+1)]=1/[(2n-1)(2(n+1)-1)]=(1/2)[1/(2n-1) -1/(2(n+1)-1)]
Tn=b1+b2+...+bn
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2(n+1)-1)]
=(1/2)[1- 1/(2n+1)]
=n/(2n+1)
n=1时,a1=S1=(c/2)×1²+(1- c/2)×1=1
n≥2时,an=Sn-S(n-1)=(c/2)×n²+(1- c/2)n-[(c/2)×(n-1)²+(1- c/2)×(n-1)]=nc-c+1
n=1时,a1=c-c+1=1,同样满足通项公式
数列{an}的通项公式为an=nc-c+1
a1、a2、a5成等比,则a2²=a1·a5
(2c-c+1)²=1·(5c-c+1)
整理,得c²-2c=0
c(c-2)=0
c=0或c=2
c=0时,an=1 a1=a2=a5,公比为1,与已知矛盾,舍去
c=2
(2)
c=2代入{an}通项公式
an=2n-2+1=2n-1
bn=1/[ana(n+1)]=1/[(2n-1)(2(n+1)-1)]=(1/2)[1/(2n-1) -1/(2(n+1)-1)]
Tn=b1+b2+...+bn
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2(n+1)-1)]
=(1/2)[1- 1/(2n+1)]
=n/(2n+1)
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