a1=3,an+1=3an/an+2,求an
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你这题若是a(n+1)=3a(n)/a(n+2),则没法求解
若是a(n+1)=3a(n)/[a(n)+2],则可以求解
由 a(n+1)=3a(n)/[a(n)+2] 可得:
1/a(n+1) = [a(n)+2]/[3a(n)] = 1/3 + (2/3)[1/a(n)]
即:1/a(n+1) - 1 = 1/3 + (2/3)[1/a(n)] - 1 = (2/3)[1/a(n)] - 2/3 = (2/3)[1/a(n) - 1]
数列 { 1 - 1/a(n) } 满足:[1 - 1/a(n+1)] / [1 - 1/a(n)] = 2/3
1 - 1/a(n) = [1 - 1/a(1)](2/3)^(n-1)
由于:1 - 1/a(1) = 1-1/3=2/3
所以:1 - 1/a(n) = (2/3)^n
1/a(n) = 1-(2/3)^n
所以:a(n) = 1/[1-(2/3)^n] = (3^n)/(3^n-2^n)
若是a(n+1)=3a(n)/[a(n)+2],则可以求解
由 a(n+1)=3a(n)/[a(n)+2] 可得:
1/a(n+1) = [a(n)+2]/[3a(n)] = 1/3 + (2/3)[1/a(n)]
即:1/a(n+1) - 1 = 1/3 + (2/3)[1/a(n)] - 1 = (2/3)[1/a(n)] - 2/3 = (2/3)[1/a(n) - 1]
数列 { 1 - 1/a(n) } 满足:[1 - 1/a(n+1)] / [1 - 1/a(n)] = 2/3
1 - 1/a(n) = [1 - 1/a(1)](2/3)^(n-1)
由于:1 - 1/a(1) = 1-1/3=2/3
所以:1 - 1/a(n) = (2/3)^n
1/a(n) = 1-(2/3)^n
所以:a(n) = 1/[1-(2/3)^n] = (3^n)/(3^n-2^n)
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