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an=3a(n-1)/[2a(n-1)+1]
1/an = [2a(n-1)+1]/[3a(n-1)]
= (2/3) + (1/3)[1/a(n-1)]
1/an - 1 = (1/3){ [1/a(n-1)] - 1 }
=>{1/an - 1} 是等比数列, q=1/3
1/an - 1 = (1/3)^(n-1) .(1/a1 - 1)
= 2.(1/3)^n
an = 1/( 1+2.(1/3)^n)
1/an = [2a(n-1)+1]/[3a(n-1)]
= (2/3) + (1/3)[1/a(n-1)]
1/an - 1 = (1/3){ [1/a(n-1)] - 1 }
=>{1/an - 1} 是等比数列, q=1/3
1/an - 1 = (1/3)^(n-1) .(1/a1 - 1)
= 2.(1/3)^n
an = 1/( 1+2.(1/3)^n)
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