Question 880182
Question 880183
<pre>
Put them in a column

  1
  4
 20
126

Subtract each number from the one below it and write it out
to the side in a new column:

  1   3
  4  16 
 20 106
126

In the 2nd column subtract each number from the one below it and 
write it out to the side in a new column:

  1   3  13  77 
  4  16  90
 20 106
126

In the 3rd column subtract each (the only) number from the one below 
it (the only one below it) and  write it out to the side in a new column:

  1   3  13  77
  4  16  90
 20 106
126

Now that we have completed the difference table,
write another 77 underneath the last number 77

  1   3  13  77
  4  16  90  <font color="red">77</font>
 20 106
126

Add the 77 and the 90, getting 167 and write that underneath 
the 90


  1   3  13  77
  4  16  90  <font color="red">77</font>
 20 106 <font color="red">167</font>
126

Add the 167 and the 106, getting 273, and write that underneath 
the 106:


  1   3  13  77
  4  16  90  <font color="red">77</font>
 20 106 <font color="red">251</font>
126 <font color="red">273</font>

Finally, add the 273 and the 126, getting 399, and write that underneath 
the 126:


  1   3  13  77
  4  16  90  <font color="red">77</font>
 20 106 <font color="red">251</font>
126 <font color="red">273</font>
<font color="red">399</font>

Answer: fifth term = 399

--------------------------------------------

There is a way to get the nth term formula:

{{{(77n^3-423n^2+748n-396)/6}}}

Substitute n=1

{{{(77n^3-423n^2+748n-396)/6}}} = {{{(77(1)^3-423(1)^2+748(1)-396)/6}}} = {{{6/6}}} = 1

Substitute n=2


{{{(77n^3-423n^2+748n-396)/6}}} = {{{(77(2)^3-423(2)^2+748(2)-396)/6}}} ={{{24/6}}} = 4

Substitute n=3

{{{(77n^3-423n^2+748n-396)/6}}} = {{{(77(3)^3-423(3)^2+748(3)-396)/6}}} = {{{120/6}}} = 20

Substitute n=4


{{{(77n^3-423n^2+748n-396)/6}}} = {{{(77(4)^3-423(4)^2+748(4)-396)/6}}} = {{{1008/6}}} = 168

Substitute n=5


{{{(77n^3-423n^2+748n-396)/6}}} = {{{(77(5)^3-423(5)^2+748(5)-396)/6}}} = {{{2394/6}}} = 399

Edwin</pre>