Question

what are the first five terms of 3n^2-1?

Answer 1

Also the second differences are 2 so c = 2 The first term is 2 so a = 2.
Using the formula, nth term = 2 + (n - 1)x1 + ½(n - 1)(n - 2)x2.
Getting rid of brackets (and noticing that ½ x 2 = 1):
nth term = 2 + n - 1 + n2 - 3n + 2.

Answer 2

Hi

Here is ur answer

Given:

  an= 3n² -1

If n=1, then

    a1 = 3 (1)² - 1

        = 3 - 1

       = 2

If n=2, then

a2= 3 (2)² -1

    = 3 ×4 -1

    = 12 - 1

    = 11

If n=3, then

a3= 3(3)² - 1

   =3 ×9 -1

  =27 -1

   = 26

If n=4, then

a4= 3 (4)² -1

   =3 ×16 -1

   = 48 - 1

   = 47

If n=5, then

a5 = 3(5)² -1

   =3 × 25 -1

   = 75 - 1

   =74

Therefore, the first five terms are 2, 11, 26, 47, 74