1, 4, 10, 20, 35, 56...

There is no one way that always works with every sequence.
But I always try a difference table first.   

List the given numbers in a column:

 1
 4
10
20
35
56

Subtract every number from the one directly below it, and write the 
result out beside each number creating a second column of numbers:

 1   3
 4   6
10  10 
20  15 
35  21
56

Do that with the second column and continue this procedure until all
the numbers in a column are the same:

 1   3   3   1
 4   6   4   1
10  10   5   1
20  15   6 
35  21
56 

Since it takes 3 columns after the first to get them all the same, 
that means that a polynomial formula for the kth term must be of degree 3.

So we assume that the nth term is of the form:



So we substitute k=1,2,3,4 and the given numbers in that and get this 
system of equations:



which reduces to 



Eliminate the D's by subtracting the 
1st equation from the 2nd,
2nd equation from the 3rd,
3rd equation from the 4th,



Eliminate the C's by subtracting the 
1st equation from the 2nd,
2nd equation from the 3rd,



Eliminate the B's by subtracting the 
1st equation from the 2nd:




Substitute in 
 




 
Substitute in 



Multiply through by LCD = 6 to clear the fractions





Substitute in



Multiply through by LCD = 6 to clear the fractions





So the formula is:


Factor out 



So 





We have formulas for all three of those sums:







So we now have:





Get an LCD inside the big parentheses:




Factor  out of the big parentheses:



Simplify the fraction in front and factor n(n+1) out of the big
parentheses:



Remove the little parentheses inside the big parentheses:



Combine terms inside:



Factor the quadratic trinomial:



Edwin