(i) tn = 4n – 3 (1 mark)


(i) tn = 4n – 3 (1 mark)

Sol. tn = 4n – 3
t1 = 4 (1) – 3 = 4 – 3 = 1
t2 = 4 (2) – 3 = 8 – 3 = 5
t3 = 4 (3) – 3 = 12 – 3 = 9
t4 = 4 (4) – 3 = 16 – 3 = 13
t5 = 4 (5) – 3 = 20 – 3 = 17

The first five terms of the sequence are 1, 5, 9, 13 and 17.

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