已知an=(2n-1)/2^(n-1),
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{(2n-1)/2^n}= 2n/2^n - 1/2^n 对于后一部分 1/2^n ,其前n项和为等比数列求和 S2 = 1/2 + 1/2^2 + 1/2^3 + …… 1/2^n = (1/2) * [1 - (1/2)^n]/(1 - 1/2) = 1 - 1/2^n 对于前一部分 2n/2^n S1 = 2*(1/2 + 2/2^2 + 3/2^3 + …… + n/2^n) 两端乘2 2S1 = 2 * [1 + 2/2 + 3/2^2 + …… + n/2^(n-1)] 两式相减,将分母方次相同的项凑在一起 2S1 - S1 = S1 = 2*{ 1 + （2/2 - 1/2）+ (3/2^2 - 2/2^2) + …… + [n/2^(n-1) - (n-1)/2^(n-1 ) - n/2^n } = 2 * [1 + 1/2 + 1/2^2 + 1/2^(n-1) - n/2^n] = 2 * { 1 * [1 - (1/2)^n]/(1 -1/2) - n/2^n} = 2 * [2 - 1/2^(n-1) - n/2^n] = 4 - 4/2^n - 2n/2^n S = S1 - S2 = 4 - 4/2^n - 2n/2^n - 1 + 1/2^n = 3 - (3 + 2n)/2^n
