Question 388333
<pre><font face = "BatangChe" color = "indigo" size = 4><b>
(0,-1) (1,-2) (2,5) (3.38) (4,115) (5,254)

List the second coordinates in a column

 
 -1
 -2
  5
 38
115
254

Subtract each number from the number just below it and write the
difference to the right of that number in a second column. These
numbers are called the first differences.

 -1    3
 -2    7
  5   33
 38   77
115  139 
254

Subtract each number in the second column from the number just below 
it and write the difference to the right of that number in a third 
column.  These numbers are called the second differences.


 -1    3    4
 -2    7   26  
  5   33   44
 38   77   62
115  139 
254

Subtract each number in the third column from the number just below 
it and write the difference to the right of that number in a fourth 
column.  These numbers are called the third differences.


 -1   -1    8   18
 -2    7   26   18
  5   33   44   18
 38   77   62
115  139 
254

We need go no further because we have reached a column in which all
the numbers are the same.  Since this column was the column of 3rd
differences, we assume the formula for the general term is a 3rd degree
polynomial equation:

{{{a[n]}}}{{{""=""}}}{{{An^3+Bn^2+Cn+D}}}

That has 4 unknown constants, so we will substitute the first
four first coordinates for n and the second coordinate for {{{a[n]}}}

For n = 0

{{{a[0]}}}{{{""=""}}}{{{A(0)^3+B(0)^2+C(0)+D}}}
{{{-1}}}{{{""=""}}}{{{A*0+B*0+C*0+D}}}
{{{-1}}}{{{""=""}}}{{{0A+0B+0C+D}}}
{{{D}}}{{{""=""}}}{{{-1}}}

Since we know the value of D as -1, we can now just substitute in

{{{a[n]}}}{{{""=""}}}{{{An^3+Bn^2+Cn-1}}}


For n = 1

{{{a[1]}}}{{{""=""}}}{{{A(1)^3+B(1)^2+C(1)-1}}}
{{{-2}}}{{{""=""}}}{{{A*1+B*1+C*1-1}}}
{{{-2}}}{{{""=""}}}{{{A+B+C-1}}}
{{{A+B+C-1}}}{{{""=""}}}{{{-2}}}
{{{A+B+C}}}{{{""=""}}}{{{-1}}}

For n = 2

{{{a[2]}}}{{{""=""}}}{{{A(2)^3+B(2)^2+C(2)-1}}}
{{{5}}}{{{""=""}}}{{{A*8+B*4+C*2-1}}}
{{{5}}}{{{""=""}}}{{{8A+4B+2C-1}}}
{{{8A+4B+2C-1}}}{{{""=""}}}{{{5}}}
{{{8A+4B+2C}}}{{{""=""}}}{{{6}}}
{{{4A+2B+C}}}{{{""=""}}}{{{3}}}

For n = 3

{{{a[3]}}}{{{""=""}}}{{{A(3)^3+B(3)^2+C(3)-1}}}
{{{38}}}{{{""=""}}}{{{A*27+B*9+C*3-1}}}
{{{38}}}{{{""=""}}}{{{27A+9B+3C-1}}}
{{{27A+9B+3C-1}}}{{{""=""}}}{{{38}}}
{{{27A+9B+3C}}}{{{""=""}}}{{{39}}}

So we have the system of 3 equations in 3 variables

{{{system(A+B+C=-1,4A+2B+C=3, 27A+9B+3C=39)}}}

Solve that system and get A=3, B=-5, C=1 and we already have D=-1,so

{{{a[n]}}}{{{""=""}}}{{{An^3+Bn^2+Cn+D}}}

becomes:

{{{a[n]}}}{{{""=""}}}{{{3n^3-5n^2+n-1}}}

Edwin</pre>