已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An}的前n项和Sn

用递推公式
a4=2a3+2^4-1=81 a3=33=2a2+2^3-1
so a2=13=2a1+2^2-1
so a1=5

an=2an-1+2^n-1
=2[2an-2+2^(n-1)-1]+2^n-1
=2^2an-2+（2^n+2^n）-(2+1)
=2^2(2an-3+2^(n-2)-1)++（2^n+2^n）-(2+1）
=2^3an-3+(2^n+2^n+2^n)-(2^2+2+1)
.......................
..........................以此类推
=2^(n-1)*a1+(n-1)*2^n-[(2^(n-2)+2^(n-3)....+2+1]
=5*2^(n-1)+(2n-2)*2^(n-1)-(1-2^(n-1))/(1-2)
=5*2^(n-1)+(2n-2)*2^(n-1)-2^(n-1)+1
=(5+2n-2-1)*2^(n-1)+1
=(2n+2)*2^(n-1)+1
=(n+1)*2^n+1

所以an=(n+1)*2^n+1

令an=bn+1 bn=(n+1)*2^n
bn前n项和为Tn
则Sn=Tn+n
Tn=2*2+3*2^2+4*2^3...........(n+1)*2^n
2*Tn=2*2^2+3*2^3..............n*2^n+(n+1)*2^(n+1)

以上两式相减得-Tn=2+2^2+2^3+。。。。。2^n-(n+1)*2^(n+1)
-Tn=2*(1-2^n)/(1-2)-(n+1)*2^(n+1)
=-n*2^(n+1)-2
so Tn=n*2^(n+1)+2
Sn=Tn+n=n*2^(n+1)+n+2

so Sn=n*2^(n+1)+n+2

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an=2a(n-1)+2^n-1,a4=2a3+2^4-1,a3=33