Find the first four terms of the sequence, whose nth term is tn = n2 – 2n

Solution:

Here, tn = n2 – 2n

∴ t1 = 12 – 2(1) = 1 – 2 = - 1

∴ t2 = 22  - 2 (2) = 4 – 4  = 0

∴ t3 = 32 – 2 (3) = 9 – 6 = 3

∴ t4 = 42 – 2(4) = 16 – 8 = 8

∴ The first four terms are – 1 , 0, 3 and 8