a(n+1) - 1/2 = [5+2a(n)]/[16 - 8a(n)] - 1/2 = [5 + 2a(n) - 8 + 4a(n)]/[16 - 8a(n)] = 3[a(n) - 1/2]/[8-4a(n)],
因a(1)=1不等于1/2,所以,由归纳法知,a(n)不等于1/2.
设,
b(n) = 1/[a(n)-1/2],a(n) = 1/b(n) + 1/2,
b(n+1) = [8 - 4/b(n) - 2]b(n)/3 = [6b(n) - 4]/3 = 2b(n) - 4/3,
b(n+1) - 4/3 = 2[b(n) - 4/3],
{b(n)-4/3}是首项为b(1)-4/3=1/[a(1)-1/2]-4/3=1/[1/2]-4/3=2/3,公比为2的等比数列.
b(n) - 4/3 = 2/3*2^(n-1) = 2^n/3, n = 1,2,...
b(n) = 4/3 + 2^n/3 = [2^n + 4]/3, n = 1,2,...
a(n) = 1/b(n) + 1/2 = 3/[2^n + 4] + 1/2, n = 1,2,...