1.由A1=1 A(n+1)=(1+1/n)An+(n+1)/2^n得
A(n+1)/(n+1)-An/n=1/2^n
設(shè)Bn=An/n B(n+1)=A(n+1)/(n+1) B1=A1=1
則B(n+1)-Bn=1/2^n
Bn-B(n-1)=1/2^(n-1)
.
B2-B1=1/2
疊加Bn-B1=1/2+1/2^2+...+1/2^(n-1)=(1/2)[1-1/2^(n-1)]/(1-1/2)=1-1/2^(n-1)
Bn=2-1/2^(n-1)
An=nBn=2n-n/2^(n-1)
設(shè)Sn=2∑n-∑n/2^(n-1)=n(n+1)-Tn
Tn=1+2/2^1+...+n/2^(n-1)
(1/2)Tn=1/2^1+2/2^2+...+n/2^n
錯位相減(1/2)Tn=1+1/2^1+1/2^2+...+1/2^(n-1)-(n+1)/2^n=(1-1/2^n)/(1-1/2)-n/2^n
=2-2/2^n-n/2^n=2-(n+2)/2^n
∴Tn=4-(n+2)/2^(n-1)
∴Sn=n^2+n+4-(n+2)/2^(n-1)
2.Sn=-An-1/2^(n-1)+2 S1=a1=-a1-1+2 a1=1/2
S(n-1)=-A(n-1)-1/2^(n-2)+2
∴An=Sn-S(n-1)=A(n-1)-An+1/2^(n-1)
∴An*2^n-A(n-1)*2^(n-1)=1
即{An*2^n}是公差為1的等差數(shù)列
An*2^n=2A1+n-1=n
∴(1) An=n/2^n
(2) Bn=(n+1)An/n=(n+1)/2^n
Tn=2/2^1+3/2^2+...+(n+1)/2^n
(1/2)Tn=2/2^2+3/2^3+...+(n+1)/2^(n+1)
錯位相減(1/2)Tn=2/2^1+1/2^2+...+1/2^n-(n+1)/2^(n+1)=1/2+(1/2)*(1-1/2^n)/(1-1/2)-(n+1)/2^(n+1)
=1/2+1-1/2^n-(n+1)/2^(n+1)
∴Tn=3-(n+1)/2^n-2/2^n=3-(n+3)/2^n
∴Tn