n数列an的前n项和Sn满足:Sn=2an-n,(1)求a1,a2,a3(2)求证:数列{an+1}是等比数列(3)bn=nan,求数列{bn}的前n项和Tn
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![n数列an的前n项和Sn满足:Sn=2an-n,(1)求a1,a2,a3(2)求证:数列{an+1}是等比数列(3)bn=nan,求数列{bn}的前n项和Tn](/uploads/image/z/11720795-59-5.jpg?t=n%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3%EF%BC%9ASn%3D2an-n%2C%281%29%E6%B1%82a1%2Ca2%2Ca3%282%29%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%2B1%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%883%EF%BC%89bn%3Dnan%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
n数列an的前n项和Sn满足:Sn=2an-n,(1)求a1,a2,a3(2)求证:数列{an+1}是等比数列(3)bn=nan,求数列{bn}的前n项和Tn
n数列an的前n项和Sn满足:Sn=2an-n,(1)求a1,a2,a3(2)求证:数列{an+1}是等比数列(3)bn=nan,求数列{bn}的
前n项和Tn
n数列an的前n项和Sn满足:Sn=2an-n,(1)求a1,a2,a3(2)求证:数列{an+1}是等比数列(3)bn=nan,求数列{bn}的前n项和Tn
(1)、S1=2a1-1=a1,a1=1
S2=a1+a2=2a2-2,a2=3
S3=a1+a2+a3=2a3-3,a3=7
(2)、因为Sn=2an-n①
所以S(n-1)=2a(n-1)-(n-1)②
①-②得:an-1-2a(n-1)=0
an+1=2[a(n-1)+1]
(an+1)/[a(n-1)+1]=2,
所以数列{an+1}是以首项1+1=2,公比为2的等比数列,
既有an+1=2×2^(n-1)=2^n,所以an=2^n-1
(3)、bn=nan=n(2^n-1)=n2^n-n
Tn=2^1-1+2×2^2-2+3×2^3-3+...+n2^n-n=2^1+2×2^2+3×2^3+...+n2^n-n(n+1)/2①
2Tn=2^2+2×2^3+3×2^4+...+n2^(n+1)-n(n+1)②
①-②得﹣Tn=2^1+2^2+2^3+.+2^n-n×2^(n+1)+n(n+1)/2
Tn=(n-1)2^(n+1)+2-n(n+1)/2