设数列{an}的前n项和为Sn,已知Sn=2n-an(n∈N+），通过计算

来源：百度知道编辑：UC知道时间：2024/06/15 11:41:26

设数列{an}的前n项和为Sn,已知Sn=2n-an(n∈N+），通过计算前四项，猜想an=?

S(1) = 2 - a(1) = a(1), a(1) = 1,

S(2) = 4 - a(2) = 1 + a(2), a(2) = 3/2,

S(3) = 6 - a(3) = 1 + 3/2 + a(3), a(3) = 7/4,

S(4) = 8 - a(4) = 1 + 3/2 + 7/4 + a(4), a(4) = 15/8

a(n) = [2^n - 1]/2^(n-1)

S(n+1) = 2(n+1) - a(n+1),
S(n) = 2n - a(n),

a(n+1) = S(n+1) - S(n) = 2 - a(n+1) + a(n),

2a(n+1) = 2 + a(n),

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

b(n) = a(n) - 2是公比为1/2的等比数列。
b(1) = a(1) - 2 = -1.
a(n) - 2 = b(n) = -(1/2)^(n-1),

a(n) = 2 - 1/2^(n-1), n = 1,2,...

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