Question

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

Answer 1

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

哇...终于码完了=.=.=累死我廖!给点辛苦分吧.