a(1)=5/6,
n>1時,
a(n+1)=a(n)/3+(1/2)^(n+1),a(2)=a(1)/3+(1/2)^2=5/18+1/4=19/36
a(n) = a(n-1)/3+(1/2)^n,
a(n)/2 = a(n-1)/6+(1/2)^(n+1),
b(n)=a(n+1)-a(n)/2 = a(n)/3-a(n-1)/6 = [a(n)-a(n-1)/2]/3 = b(n-1)/3
{b(n)}是首項為b(1)=a(2)-a(1)/2=19/36-5/12=1/9,公比為1/3的等比數(shù)列.
a(n+1)-a(n)/2=b(n)=(1/9)(1/3)^(n-1)=(1/3)^(n+1),
3^(n+1)a(n+1) = 1 + (3/2)3^na(n),
3^(n+1)a(n+1)+2=(3/2)[3^na(n)+2],
{3^na(n)+2}是首項為3a(1)+2=3*5/6+2=9/2,公比為3/2的等比數(shù)列.
3^na(n)+2=(9/2)(3/2)^(n-1)=3*(3/2)^n,
a(n) = [3*(3/2)^n - 2]/3^n = 3/2^n - 2/3^n