Question 1156464
<br>
The differences between terms form a logical pattern, so we can find a polynomial that produces the given sequence.<br>
Here is a display showing that the sequence produces a constant third difference:<br><pre>
    0     1     4    10    20    35    56     <-- original sequence
       1     3     6    10    15    21        <-- first differences
          2     3     4     5     6           <-- second differences
             1     1     1     1              <-- third differences</pre><br>
The constant third difference tells us the sequence can be produced by a polynomial of degree 3:  {{{t(n) = an^3+bn^2+cn+d}}}<br>
If you know some basic differential calculus, that is because the n-th derivative of a polynomial of degree n is a constant.<br>
At this point, we could make a system of four equations in the four variables a, b, c, and d using the first four terms of the sequence:<br>
t(1):  {{{a+b+c+d=0}}}
t(2):  {{{8a+4b+2c+d=1}}}
t(3):  {{{27a+9b+3c+d=4}}}
t(4):  {{{64a+16b+4c+d=10}}}<br>
Finishing the problem by that path is a good exercise.  It would be well worth your time and effort to do that and see that you get the right polynomial.<br>
However, this example is a good one to show that often there is a faster and easier way to find the polynomial.<br>
Let's go back to that earlier mention about derivatives of a polynomial.<br>
Given a polynomial of degree 3 with leading coefficient 1, the third derivative is 3! = 6:<br>
function: x^3+....
1st derivative: 3x^2+...
2nd derivative: 6x+...
3rd derivative: 6<br>
Since in this problem the constant 3rd difference is 1, we know that the leading coefficient of the polynomial that produces the sequence is 1/6.<br>
So one thing we could do is compare the terms of the given sequence to the terms of the sequence {{{t(x) = (1/6)x^3}}}.<br>
But that is clearly going to have us working with fractions, which always makes our calculations slower and more difficult.<br>
So what we can do now is multiply every term in the given sequence by 6; that will give us a constant 3rd difference of 1.<br>
And now we can compare the terms of the new sequence to the terms of the sequence {{{t(x) = x^3}}}.<br><pre>
    term #   6 times given term  x^3   difference
  -------------------------------------------------
      1                   0       1      -1
      2                   6       8      -2
      3                  24       27     -3
      4                  60       64     -4
      5                 120       125    -5
      6                 210       216    -6
      7                 336       343    -7</pre><br>
This analysis shows us that the sequence of 6 times the given terms is given by the formula {{{x^3-x}}}.<br>
And so the polynomial formula for the given sequence is<br>
{{{t(x) = (1/6)(x^3-x) = (x(x^2-1))/6}}}<br>
CHECK:
t(1): {{{(1*0)/6 = 0}}}
t(2): {{{(2*3)/6 = 1}}}
t(3): {{{(3*8)/6 = 4}}}
t(4): {{{(4*15)/6 = 10}}}
t(5): {{{(5*24)/6 = 20}}}
t(6): {{{(6*35)/6 = 35}}}
t(7): {{{(7*48)/6 = 56}}}<br>