Question 853296
<pre>
terms     1st diffs   2nd diffs  3rd diffs
 -14         -1           3         -49
 -15          2         -46
 -12        -44
 -56

Since it takes 3 differences to get to a column of
differences with all the same number, we assume
a polynomial solution of degree 3:

So we assume the general term:

a<sub>n</sub> = An³+Bn²+Cn+D

a<sub>1</sub> = -14 = A(1)³+B(1)²+C(1)+D

A+B+C+D = -14

-------------------------------

a<sub>2</sub> = -15 = A(2)³+B(2)²+C(2)+D

8A+4B+2C+D = -15

------------------------------- 

a<sub>3</sub> = -12 = A(3)³+B(3)²+C(3)+D

27A+9B+3C+D = -12

-------------------------------

a<sub>4</sub> = -56 = A(4)³+B(4)²+C(4)+D

64A+16B+4C+D = -56

-------------------------------

Solve this system of equations:

  A+  B+ C+D = -14
 8A+ 4B+2C+D = -15
27A+ 9B+3C+D = -12
64A+16B+4C+D = -56

Get:
A=-8.5, B=53, C=-100.5, D=42

General term:

a<sub>n</sub> = -8.5n³+53n²-100.5n+42

-14, -15, -12, -56, -198, -489, -980, -1722, -2766, -4163, -5964, ...

Edwin</pre>