数列{an}满足：a1=2,a2=1,an>0,(an^2-an-1^2)/(an-1^2)=(an+1^2-an^2)
2个回答

[an²-a(n-1)²]/[a(n-1)²]=[a(n+1)²-an²]/[a(n+1)²]

an²/a(n-1)² -1=1 -an²/a(n+1)²

an²/a(n-1)²+an²/a(n+1)²=2

1/a(n+1)²+1/a(n-1)²=2/an²

数列{1/an²}是等差数列.

1/a2²-1/a1²=1/1²-1/2²=1-1/4=3/4

1/a1²=1/2²=1/4

数列{1/an²}是以1/4为首项,3/4为公差的等差数列.

1/a6²=1/4 +(3/4)(6-1)=4

an>0 a6>0

1/a6=1/2

a6=2

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