a(n+2)-a(n+1)=(1/3)[a(n+1)-a(n)],
{a(n+1)-a(n)}是首项为a(2)-a(1)=7/9 - 1/3 = 4/9,公比为(1/3)的等比数列.
a(n+1)-a(n) = (4/9)(1/3)^(n-1) = 4/3^(n+1),
a(n+1)3^(n+1) = 3a(n)3^(n) + 4,
2+a(n+1)3^(n+1) = 3[2 + a(n)3^n]
{2+a(n)3^(n)}是首项为2+3a(1)=2+1=3,公比为3的等比数列.
2+a(n)3^n = 3*3^(n-1)=3^n,
a(n) = 1 - 2/3^n
na(n) = n - 2n/3^n
s(n) = 1-2/3 + 2-2*2/3^2 + ... +(n-1)-2(n-1)/3^(n-1) + n-2n/3^n
=1+2+...+(n-1)+n - 2[1/3 + 2/3^2 + ... + (n-1)/3^(n-1) + n/3^n]
=n(n+1)/2 - 2t(n).
t(n) = 1/3 + 2/3^2 + ... + (n-1)/3^(n-1) + n/3^n
3t(n) = 1 + 2/3 + ... + (n-1)/3^(n-2) + n/3^(n-1)
2t(n) = 3t(n) - t(n) = 1 + 1/3 + ... + 1/3^(n-2) + 1/3^(n-1) - n/3^n
=[1-1/3^n]/(1-1/3) - n/3^n
= (3/2)[1-1/3^n] - n/3^n
s(n)=n(n+1)/2 - 2t(n)
=n(n+1)/2 -(3/2)[1-1/3^n] + n/3^n