(iv) tn = n2 – 2n (1 mark)

(iv) t_n = n² – 2n (1 mark)

Sol. t_n = n² – 2n

∴ t₁ = 1 – 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

∴ t₅ = 52 – 2 (5) = 25 – 10 = 15

∴ The first five terms of the sequence are – 1, 0, 3, 8 and 15.