数列{an}的前n项和记为Sn,n,an,Sn成等差数列(n∈N*),证明:(Ⅰ...
数列{an}的前n项和记为Sn,n,an,Sn成等差数列(n∈N*),证明:(Ⅰ)数列{an 1}为等比数列
(Ⅱ)求{an}的通项公式
n,an,Sn成等差数列,所以n Sn=2an ,即 Sn=2an - n ,
an 1 = Sn 1 - Sn = 2an 1 - n-1 - 2an n = 2an 1 - 2an -1
化简就是an 1 = 2an 1
an 1 1 = 2an 2 =2(an 1)
( an 1 1)/(an 1)=2
n =1 时,1 S1=2a1;a1=1
数列{an 1}为等比数列
问:数学数列An的n项和记为Sn,已知A1=1,An+1=(n+2/n)Sn(n=123456......)证明{Sn/n}是等比数列
答:根据已知条件,有 S_(n+1)=S_n+A_(n+1)=S_n+[(n+2)/n]S_n=S_n2[(n+1)/n] S_(n+1)/(n+1)=2(S_n/...详情>>
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