Question 1184909
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The response from the other tutor shows that the second differences for the given sequence are constant:<br><pre>
    1   7  17  31  49  71
      6  10  14  18  22
        4   4   4   4</pre>
The constant second differences mean the sequence can be generated by a second degree polynomial of the form<br>
{{{t(n)=an^2+bn+c}}}<br>
In his solution, he uses the general form of the quadratic with the first three terms of the sequence to get three equations in the coefficients a, b, and c, which he then solves.<br>
n=1: a+b+c=1
n=2: 4a+2b+c=7
n=3: 9a+3b+c=17
...
a=2; b=0; c=-1<br>
Here is another way to get from the constant second differences to the polynomial.<br>
The constant difference of 4 tells us that the sequence can be generated by a second degree polynomial.  But it tells us more: it tells us that the leading coefficient a is 4/(2!) = 4/2 = 2.<br>
So the polynomial is of the form<br>
{{{t(n)=2n^2+bn+c}}}<br>
We can find the "bn+c" part of the polynomial by comparing the terms of the sequence to the value of 2n^2.<br><pre>
     n  t(n) 2n^2  difference
                  t(n)-2n^2=bn+c
   -------------------------------
     1   1     2     -1
     2   7     8     -1
     3  17    18     -1</pre>
We see that the difference between t(n) and 2n^2 is the constant -1; that means the polynomial that generates the sequence is<br>
{{{t(n)=2n^2-1}}}<br>