Question 1012259
<pre>
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:

{{{a[k]=A*k^3+B*k^2+C*k+D}}}

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

{{{system(
1=A*(1)^3+B*(1)^2+C*(1)+D,
4=A*(2)^3+B*(2)^2+C*(2)+D,
10=A*(3)^3+B*(3)^2+C*(3)+D,
20=A*(4)^3+B*(4)^2+C*(4)+D)}}}

which reduces to 

{{{system(
A+B+C+D=1,
8A+4B+2C+D=4,
27A+9B+3C+D=10,
64A+16B+4C+D=20)}}}

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

{{{system(
7A+3B+C=3,
19A+5B+C=6,
37A+7B+C=10)}}}

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

{{{system(
12A+2B=3,
18A+2B=4)}}}

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

{{{6A=1}}}
{{{A=1/6}}}

Substitute in 
{{{12A+2B=3}}} 
{{{12(1/6)+2B=3}}}
{{{2+2B=3}}}
{{{2B=1}}}
{{{B=1/2}}}
 
Substitute in 
{{{7A+3B+C=3}}}
{{{7(1/6)+3(1/2)+C=3}}}
{{{7/6+3/2+C=3}}}
Multiply through by LCD = 6 to clear the fractions
{{{7+9+6C=18}}}
{{{16+6C=18}}}
{{{6C=2}}}
{{{C=1/3}}}

Substitute in

{{{A+B+C+D=1}}}
{{{1/6+1/2+1/3+D=1}}}
Multiply through by LCD = 6 to clear the fractions
{{{1+3+2+6D=6}}}
{{{6+6D=6}}}
{{{6D=0}}}
{{{D=0}}}

So the formula is:

{{{a[k]=expr(1/6)k^3+expr(1/2)k^2+expr(1/3)k}}}
Factor out {{{1/6}}}

{{{a[k]=expr(1/6)(k^3+3k^2+2k)}}}

So 

{{{sum((a[k]),k=1,n)}}}{{{""=""}}}{{{sum(( expr(1/6)(k^3+3k^2+2k)),k=1,n)}}}{{{""=""}}}{{{expr(1/6)sum( (k^3+3k^2+2k),k=1,n)}}}{{{""=""}}}

{{{expr(1/6)(sum( (k^3),k=1,n)+sum( (3k^2),k=1,n)+sum( (2k^""),k=1,n))}}}{{{""=""}}}{{{expr(1/6)(sum( (k^3),k=1,n)+3*sum( (k^2),k=1,n)+2*sum( (k^""),k=1,n))}}}{{{""=""}}}

We have formulas for all three of those sums:

{{{sum((k^3),k=1,n)=expr(1/4)n^2(n+1)^2}}}

{{{sum((k^2),k=1,n)=expr(1/6)n(n+1)(2n+1)}}}

{{{sum((k),k=1,n)=expr(1/2)n(n+1)}}}

So we now have:

{{{expr(1/6)(expr(1/4)n^2(n+1)^2+3*expr(1/6)n(n+1)(2n+1)+2*expr(1/2)n(n+1)^"")}}}{{{""=""}}}

{{{expr(1/6)(expr(1/4)n^2(n+1)^2+expr(1/2)n(n+1)(2n+1)^""+n(n+1)^"")^""}}}{{{""=""}}}

Get an LCD inside the big parentheses:

{{{expr(1/6)(expr(1/4)n^2(n+1)^2+expr(2/4)n(n+1)(2n+1)^""+expr(4/4)n(n+1)^"")^""}}}{{{""=""}}}


Factor {{{1/4}}} out of the big parentheses:

{{{expr(1/6)expr(1/4)(n^2(n+1)^2+2n(n+1)(2n+1)^""+4n(n+1)^"")^""}}}{{{""=""}}}

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

{{{expr(1/24)n(n+1)(n(n+1)+2(2n+1)^""+4^"")^""}}}{{{""=""}}}

Remove the little parentheses inside the big parentheses:

{{{expr(1/24)n(n+1)(n^2+n+4n+2^""+4^"")^""}}}{{{""=""}}}

Combine terms inside:

{{{expr(1/24)n(n+1)(n^2+5n+6^"")^""}}}{{{""=""}}}

Factor the quadratic trinomial:

{{{expr(1/24)n(n+1)(n+2)(n+3)}}}

Edwin</pre>