0, we can find explicit formulae
for the kth term, bk, and the (k-1)th term, bk-1. 3. Given the sequence 1, 1, 2, 6, 24, 120, ... , n!, ... for n0, we can find explicit formulae
for the kth term, bk, and the (k-1)th term, bk-1.
bk = k! and bk-1 = (k - 1)!.
Hence k · bk-1 = k · (k-1)! = k! = bk.
Thus this sequence satisfies the recurrence relation bk
= k · bk-1 for all integers k1.