Question

在数列{an}中，a1=，an+1=(1+1/n)an+n+1/2^n

在数列{an}中，a1=1，an+1=(1+1/n)an+(n+1)/2^n
求bn=an/n，求an的前n项和Sn

Answer 1

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

s(n)=2n(n+1)/2 - t(n)
=n(n+1) - 4 + (n+2)/2^(n-1)