设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+1)+1,n∈N,且a1,a2+5,a3成等差数列

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设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+1)+1,n∈N,且a1,a2+5,a3成等差数列
求数列an的通项公式.
2Sn=a(n+1)-2^(n+1)+1 2S(n-1)=an-2^(n)+1 2an=2Sn-2S（n-1）
2an = a(n+1) -an -2^n
a(n+1) = 3an+2^n
a(n+1) + 2^(n+1) = 3(an + 2^n) 这部怎么来的?

待定系数设a(n+1)+k^(n+1)=3(an+k^n) k=2

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