a1=1.an+1=2an+2^n.bn=an/2^n-1.证明bn是等差数列、求数列的前n项和sn?

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a(n+1)=2an+2^n两边÷2^n
a(n+1)/2^n=an/2^(n-1)+1
bn=an/2^n-1 bn+1=a(n+1)/2^n
bn+1-bn=1 bn是等差
Sn=1+2*2+.+n*2^(n-1)
2Sn=2+2*2^2+...+n*2^n
Sn-2Sn=1+2+.+2^(n-1)-n*2^n
Sn=(n-1)*2^n+1

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