Question 852459
<pre>
Since we have 4 terms, we will assume a third degree 
polynomial for the nth term, because a third degree polynomial
has four coefficients.

That is, we assume a virtual function of the form

Y = A*N^3+B*N^2+C*N+D

Substitute these for (N,Y) (1,22), (2,59), (3,120), and (5,358)

    A*1^3+B*1^2+C*1+D = 22
    A*2^3+B*2^2+C*2+D = 22
    A*3^3+B*3^2+C*3+D = 22
    A*5^3+B*5^2+C*5+D = 358

And solve the augmented matrix:

{{{(matrix(4,6,
    1^3, 1^2, 1^1, 1, "|", 22,
    2^3, 2^2, 2^1, 1, "|", 59,
    3^3, 3^2, 3^1, 1, "|", 120,
    5^3, 5^2, 5^1, 1, "|", 358))}}}

We solve that matrix and get 

A = 1.83333333
B = 1
C = 21.16666667
D = -2

Then we substitute X=4 in

Y = A*N^3+B*N^2+C*N+D

and get 216.

Edwin</pre>