an+1=2an+2的n次方-1
两边同除以 2^(n+2)
an+1/ 2^(n+2) = (2an+2^(n-1))/2^(n+2) = an/2^(n+1) + 1/8
an+1/ 2^(n+2) - an/2^(n+1) = 1/8
所以an/2^(n+1) 为等差数列，即an-1/2^n 为等差数列
设bn = an/2^(n+1)
b1 = a1/2^2 =1/4
bn = b1+1/8 *(n-1) = 1/4 +1/8 n - 1/8 = 1/8 n + 1/8
an = 2^(n+1) * bn = 2^(n+1) * (1/8 n + 1/8)
=n 2^(n-2) + 2^(n-2)
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