数列{an}首相a1≠0,前n相的和为sn,满足Sn+1=2sn+a1,那么2an/sn等于多少,n趋近正无穷
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数列{an}首相a1≠0,前n相的和为sn,满足Sn+1=2sn+a1,那么2an/sn等于多少,n趋近正无穷
![数列{an}首相a1≠0,前n相的和为sn,满足Sn+1=2sn+a1,那么2an/sn等于多少,n趋近正无穷](/uploads/image/z/6125952-48-2.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E9%A6%96%E7%9B%B8a1%E2%89%A00%2C%E5%89%8Dn%E7%9B%B8%E7%9A%84%E5%92%8C%E4%B8%BAsn%2C%E6%BB%A1%E8%B6%B3Sn%2B1%3D2sn%2Ba1%2C%E9%82%A3%E4%B9%882an%2Fsn%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91%2Cn%E8%B6%8B%E8%BF%91%E6%AD%A3%E6%97%A0%E7%A9%B7)
Sn+1=Sn+a(n+1)=2Sn+a1
Sn=a(n+1)-a1
Sn-1=an-a1
an=Sn-Sn-1=a(n+1)-an
a(n+1)=2an
a(n+1)/an=2,为定值.
数列{an}是以a1为首项,2为公比的等比数列.
an=a1×2^(n-1) 2an=a1×2ⁿ
2an/Sn=a1×2ⁿ/[a1×(2ⁿ-1)/(2-1)]=2ⁿ/(2ⁿ-1)=1/(1- 1/2ⁿ)
当n->+∞时,1/2ⁿ->0 1-1/2ⁿ->1 2an/Sn->1
lim(2an/Sn)=1
Sn=a(n+1)-a1
Sn-1=an-a1
an=Sn-Sn-1=a(n+1)-an
a(n+1)=2an
a(n+1)/an=2,为定值.
数列{an}是以a1为首项,2为公比的等比数列.
an=a1×2^(n-1) 2an=a1×2ⁿ
2an/Sn=a1×2ⁿ/[a1×(2ⁿ-1)/(2-1)]=2ⁿ/(2ⁿ-1)=1/(1- 1/2ⁿ)
当n->+∞时,1/2ⁿ->0 1-1/2ⁿ->1 2an/Sn->1
lim(2an/Sn)=1
数列{an}首相a1≠0,前n相的和为sn,满足Sn+1=2sn+a1,那么2an/sn等于多少,n趋近正无穷
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