(iv) tn = n2 – 2n (1 mark)


(iv) tn = n2 – 2n (1 mark)
Sol. tn = 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.

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