令A(n)=an+2^n
上面就是A(n+1)=3 A(n)
数列A(n)是等比数列
而A(1)=a1+2^1=a1+2=3
所以A(n)=A(1)*3^(n-1)=3^n
也就是an+2^n=3^n
设数列{an}的前n项和为sn,满足2sn=a(n+1)-2^(n+1)+1,n属于n*.且a1,a2+5,a3成等差数
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