List the first 4 terms of the sequence 3n² + 1

4, 13, 28, 49

Step-by-step explanation:

1st term = 3 × (1)² + 1 = 4

2nd term = 3 × (2)² + 1 = 13

3rd term = 3 × (3)² + 1 = 28

4th term = 3 × (4)² + 1 = 49

The first for terms of the series can be found by substituting the n for the term it is:

= 3(1)² + 1 , 3(2)² + 1 , 3(3)² + 1 ,3(4)² + 1

= 4, 13, 28, 49

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation: