Question

在数列an中，a1=1,an+1=(1+1/n)an+n+1/2^n 1.设bn=an/n,求数列bn的通项公式 2.求数列an的前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 = a(n)/n + 1/2^(n-1) - 1/2^n,
a(n+1)/(n+1) + 1/2^n = a(n)/n + 1/2^(n-1),
{a(n)/n + 1/2^(n-1)}是首项为a(1)+1=2,的常数数列。
a(n)/n + 1/2^(n-1) = 2,
b(n) = a(n)/n = 2- 1/2^(n-1) = [2^n - 1]/2^(n-1),

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 + 1/2 + ... + 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)