在数列an中,a1=1,an+1=(1+1/n)an+n+1/2^n 且bn=an／n,求an前n项和sn

Question

在数列an中,a1=1,an+1=(1+1/n)an+n+1/2^n 且bn=an／n,求an前n项和sn

Answer 1

a(n+1)=(n+1)a(n)/n + (n+1)/2^n,
a(n+1)/(n+1)=a(n)/n + 1/2^n,
b(n+1)=b(n) + 1/2^n,b(1)=a(1)/1=1.
2^nb(n+1)=2*[2^(n-1)b(n)] + 1,
2^nb(n+1) + 1 = 2[2^(n-1)b(n) + 1],
{2^(n-1)b(n)+1}是首项为b(1)+1=2,公比为2的等比数列.
2^(n-1)b(n)+1=2^n,
b(n)=[2^n-1]/2^(n-1)=a(n)/n,
a(n)=n[2^n-1]/2^(n-1)=2n - n/2^(n-1).
s(n)=n(n+1)-[1/1 + 2/2 + 3/2^2 + ...+ (n-1)/2^(n-2) + n/2^(n-1)]=n(n+1)-t(n)
t(n)=1/1 + 2/2 + 3/2^2 + ...+ (n-1)/2^(n-2) + n/2^(n-1)
2t(n)=2/1 + 2/1 + 3/2 + ...+ (n-1)/2^(n-3) + n/2^(n-2),
t(n)=2t(n)-t(n)=2/1 + 1/1 + 1/2 + ...+ 1/2^(n-2) - n/2^(n-1)=2+[1-1/2^(n-1)]/[1-1/2] - n/2^(n-1)
=2 + 2[1-1/2^(n-1)] - n/2^(n-1)
=4 -(n+2)/2^(n-1),
s(n) = n(n+1) - t(n) = n(n+1) - 4 + (n+2)/2^(n-1)