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.