a(n+1) = a(n)/[4a(n)+1],
若a(n+1)=0,則a(n)=0, ..., a(1)=0,與a(1)=1矛盾.因此,a(n)不為0.
1/a(n+1) = [4a(n)+1]/a(n) = 1/a(n) + 4,
{b(n) = 1/a(n)}是首項(xiàng)為b(1)=1/a(1)=1,公差為4的等差數(shù)列.
b(n) = 1 + 4(n-1) = 4n-3.
4n-3 = b(n) = 1/a(n),
a(n) = 1/(4n-3).
a(k)=1/(4k-3)
a(k+1) = 1/(4k+4-3) = 1/(4k+1),
a(k)a(k+1)=1/[(4k-3)(4k+1)] = 1/[16k^2 - 12k + 4k - 3] = 1/[16k^2 - 8k - 3]
= 1/[4(4k^2-2k) - 3]
= a(4k^2 - 2k)
a(k)a(k+1)仍是{a(n)}中的項(xiàng),是第（4k^2-2k）項(xiàng)