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已知数列{an}中,a1=4,an=3a(n-1)-2(n大于等于2)
(1)求证:数列{an-1}是等比数列,并求数列{an}的通项
(2)证明:1/a1+1/a2+...+1/an<1/2
人气:284 ℃ 时间:2020-04-18 12:38:22
解答
(1)an=3a(n-1)-2an-1 = 3(a(n-1) -1)(an-1)/(a(n-1) -1)=3(an-1)/(a1-1) =3^(n-1)an = 1+ 3^n(2)1/an = 1/(1+3^n) < 1/3^n1/a1+1/a2+..1/an < 1/3 +1/3^2 +...+1/3^n= (1/3)(1- (1/3)^n )/ (1-1/3)= (1/2)(1- (1/3)^...
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