Sn=4-an-1/(2^(n-2)),求an

用S(n+1) - Sn应该就可以得出a(n+1)与an的关系
由题意Sn=4-an-〔1/2^（n-2）〕
得S(n+1) = 4 - a(n+1) - [1/2^(n-1)]
所以S(n+1) - Sn = {4 - a(n+1) - [1/2^(n-1)]} - {4-an-〔1/2^（n-2）〕 } = -a(n+1) - 1/ 2^(n-1) + an + 1/2^(n-2)
而S(n+1) - Sn = a(n+1)
所以a(n+1) =
-a(n+1) - 1/ 2^(n-1) + an + 1/2^(n-2)
所以a(n+1) = (1/2) an + 1/2^(n-1) - 1/2^n
————————
所以an = (1/2) a(n-1) + 1/2^(n-2) - 1/2^(n-1)
即(1/2)an = 1/2^2 a(n-1) + 1/2^(n-1) - 1/2^n
所以a(n+1) = 1/2^2 a(n-1) + (1/2^(n-1) - 1/2^n)*2
所以a(n+1) = 1/2^(n-1) a1 + (1/2^(n-1) - 1/2^n)*(n-1)
由因为s1 = 4-a1-1/2^(-1)
所以a1 = 1
所以a(n+1) = 1/2^(n-1) + (1/2^(n-1) - 1/2^n)*(n-1)
即an = 1/2^(n-2) + (1/2^(n-2) - 1/2^(n-1))*(n-2)
= n/(2^(n-1))