Find the first four terms of the sequence whose nth terms are given by (iii) an = 2n2 - 6

Find the first four terms of the sequence whose nth terms are given by
(iii) an = 2n2 - 6
Solution :
an = 2n2 - 6
n = 2
=  2(2)2 - 6
a2  = 2
an = 2n2 - 6
n = 1
=  2(1)2 - 6
a1  = -4

an = 2n2 - 6
n = 3
=  2(3)2 - 6
a3  = 12
an = 2n2 - 6
n = 4
=  2(4)2 - 6
a4  = 26

Hence the four terms are -4, 2, 12, 26.

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