n   tn    D1  D2
---------------   
1   2.5  3.5  4
2   6.0  7.5  4
3  13.5 11.5
4  25.0

Since it took 2 differences to get to a column that only
had the same number in it, we can assume a 2nd degree
polynomial function for tn of the form:

tn =  An² + Bn + C

Substituting

t1 =  A(1)² + B(1) + C
2.5 = A(1) + B(1) + C
2.5 = A + B + C

t2 =  A(2)² + B(2) + C
6.0 = A(4) + B(2) + C
6.0 = 4A + 2B + C

t3 =  A(3)² + B(3) + C
13.5 = A(9) + B(3) + C
13.5 = 9A + 3B + C

So we have this system

  2.5 = A +  B + C
 6.0 = 4A + 2B + C
13.5 = 9A + 3B + C

or maybe you would write it this way:

 A +  B + C =  2.5
4A + 2B + C =  6.0
9A + 3B + C = 13.5

Solve that system of 3 equations in three
unknowns and get:

A = 2, B = -2.5, C = 3

So the function 

tn =  An² + Bn + C

becomes

tn =  2n² - 2.5n + 3

Edwin