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

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