Question 1100724
See #	Question 1100724 Answer by MathLover1

take the differences between the numbers to get:
9, 15, 21, 27

take the second differences by taking the differences of the numbers above:
6, 6, 6

all the second differences are equal to six, which means the equation is quadratic: y=ax^2+bx+c

we can find the a, b, & c values by pluggin in the points (1,3), (2,12), & (3,27) and solving for a,b,& c by elimination:

1a+1b+1c=3
4a+2b+1c=12
9a+3b+1c=27

1a+1b+1c=3
3a+1b+0c=9
8a+2b+0c=24


1a+1b+1c=3
3a+1b+0c=9
2a+0b+0c=6

2a=6
a=3

3a+b=9
b=0

a+b+c=3
c=0

final equation of nth term is: {{{3n^2+0n+0 = 3n^2}}} or

{{{ 3n^2}}}