在数列{an}中,若a₁=1,a₂=4 且an^2=a(n+1)×(a﹙n+1﹚-2^2×an),求通项公式
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在数列{an}中,若a₁=1,a₂=4 且an^2=a(n+1)×(a﹙n+1﹚-2^2×an),求通项公式
在数列{an}中,若a₁=1,a₂=4 且an^2=a(n+1)×(a﹙n+1﹚-2^2×an),求通项公式
在数列{an}中,若a₁=1,a₂=4 且an^2=a(n+1)×(a﹙n+1﹚-2^2×an),求通项公式
an^2=a(n+1)[a(n+1)-2^2an]
an/a(n+1)=a(n+1)/an-4
设bn=a(n+1)/an,b1=b2/b1=4
1/bn=bn-4
bn^2-4bn-1=0
bn=2±√5
a(n+1)/an=bn=2±√5
an=a1(2±√5)^(n-1)
=(2±√5)^(n-1)
将a1、a2代入an^2=a(n+1)[a(n+1)-2^2an]不成立,题目有误.