已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n属于N*),且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数
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![已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n属于N*),且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数](/uploads/image/z/1389355-43-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D1%2Can%2B1%3D2Sn%2B1%2C%EF%BC%88n%E5%B1%9E%E4%BA%8EN%2A%EF%BC%89%2C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Bbn%7D%E4%B8%ADbn%26gt%3B0%28n%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D1%2Can%2B1%3D2Sn%2B1%2C%EF%BC%88n%E5%B1%9E%E4%BA%8EN%2A%EF%BC%89%2C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Bbn%7D%E4%B8%ADbn%3E0%28n%E5%B1%9E%E4%BA%8EN%2A%29%2C%E4%B8%94b1%2Bb2%2Bb3%3D15%2C%E5%8F%88a1%2Bb1%E3%80%81a2%2Bb2%E3%80%81a3%2Bb3%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0)
已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n属于N*),且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数
已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n
已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n属于N*),且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数列.1)求数列{an}, {bn}的通项公式,2),求数列{an·bn}的前n项和Tn.
已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1,(n属于N*),等差数列{bn}中bn>0(n属于N*),且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数
1) an+1 = 2Sn +1 [1]
an+2 = 2Sn+1 +1 [2]
[2]-[1]:an+2 - an+1 = 2*[Sn+1 -Sn]=2*an+1
an+2=3an+1
所以 {an}是公比为3的等比数列,首项是1,an=a1*3^(n-1)=3^(n-1)
由上面的通项公式得到:a2=3,a3=9
因为{bn}为等差数列,且b1+b2+b3=15,所以b2=5,b1+b3=10
a1+b1=1+b1=1+10-b3=11-b3
a2+b2=3+5=8
a3+b3=9+b3
a1+b1,a2+b2,a3+b3成等比数列,所以 8*8=(11-b3)*(9+b3)
解此方程,得到:b3=7,所以b1=10-7=3
bn=3+(n-1)X2= 2n+1
(2)错位相减,实在不想写了.楼主应该会