数列的递推和求和

Question

如题所述

Answer 1

b1=a1/1=1/1=1

a(n+1)=(1+1/n)an+(n+1)/2^n
a(n+1)=(n+1)/n*an+(n+1)/2^n
a(n+1)/(n+1)=an/n+1/2^n
b(n+1)=bn+1/2^n
bn=b(n-1)+1/2^(n-1)
bn-b(n-1)=1/2^(n-1)

bn-b(n-1)=1/2^(n-1)
..........
b3-b2=1/2^2
b2-b1=1/2^1
以上等式相加得
bn-b1=1/2^1+1/2^2+.......+1/2^(n-1)
bn-b1=1/2*[1-1/2^(n-1)]/(1-1/2)
bn-b1=1-1/2^(n-1)
bn-1=1-1/2^(n-1)
bn=2-1/2^(n-1)

bn=an/n
an=n*bn
=n*[2-1/2^(n-1)]
=2n-n/2^(n-1)

Sn=a1+a2+.........+an
Sn=2-1/2^0+2*2-2/2^1+.........+2n-n/2^(n-1)
Sn=2*(1+2+........+n)-(1/2^0+2/2^1+.......+n/2^(n-1)
Sn=n(n+1)-(1/2^0+2/2^1+.......+n/2^(n-1)...............1
Sn/2=n(n+1)/2-(1/2^1+2/2^2+.......+n/2^n)...........2
1式-2式得
Sn/2=n(n+1)/2-[1/2^0+1/2^1+.....1/2^(n-1)]+n/2^n
Sn/2=n(n+1)/2-[1-(1/2)^n]/(1/2)+n/2^n
Sn=n(n+1)-2[1-(1/2)^n]/(1/2)+2n/2^n
Sn=n(n+1)-4[1-(1/2)^n]+2n/2^n
Sn=n(n+1)-4+4*(1/2)^n+2n/2^n
Sn=n(n+1)-4+4/2^n+2n/2^n
Sn=n^2+n-4+(n+2)/2^(n-1)