数列｛an｝中,a1=1,an＋1=2an+2^n(1)设bn=an/2^n-1.证明数列{bn}是等差数列（2）求数列{an}的前n项和sn
bn = an/2^(n-1)
b = a/2^(n-2)
bn - b
= an/2^(n-1) - a/2^(n-2)
= (an - 2a )/2^(n-1)
把 已知条件 a = 2an+2^n 即 an = 2a + 2^(n-1) 代入上式
bn - b
= 2^(n-1)/2^(n-1)
= 1
因此 bn 是等差数列
b1 = a1/2^(1-1) = 1/1 = 1
bn = n
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an/2^(n-1) = n
所以
an = n * 2^(n-1)
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Sn = a1 + a2 + a3 + …… + a + an
= 1 + 2*2 + 3*2^2 + …… + (n-1)*2^(n-2) + n * 2^(n-1)
2Sn = 2 + 2*2^2 + 3*2^3 + …… + (n-1)*2^(n-1) + n * 2^n
两式子相减,把 2的乘方相同的相合并在一起
2Sn - Sn = Sn
= -1 + (1-2)*2 + (2-3)*2^2 + (3-4)*2^3 + …… [(n-1) -n]*2^(n-1) + n*2^n
= n*2^n - [ 1 + 2 + 2^2 + …… 2^(n-1)]
= n*2^n - 1*(2^n -1)/(2-1)
= n * 2^n - 2^n + 1
= (n-1)*2^n + 1
这个解析的bn - b
= an/2^(n-1) - a/2^(n-2)
= (an - 2a )/2^(n-1) 我就是化不起来,怎么通分的啊.、