Question 174497
<pre><font size = 4 color = "indigo"><b>
{{{matrix(1,15,
(n-0)*1, "+", (n-1)*2, "+", (n-2)*3, "+", (n-3)*4, "+", "...", "+", 3*(n-2),"+",2*(n-1),"+",1*n )}}} =

{{{sum( matrix(1,4, "", "[", n-(k-1),"]*k" ),k=1,n)}}} =

{{{sum( matrix(1,2, "", (n-k+1)k ),k=1,n)}}} =

{{{sum( matrix(1,2, "", nk-k^2+k ),k=1,n)}}} =

{{{sum( matrix(1,2, "", nk ),k=1,n) - sum( matrix(1,2, "",k^2 ),k=1,n)+sum( matrix(1,2, "", k ),k=1,n)}}}

{{{n*sum( matrix(1,2, "", k ),k=1,n) - sum( matrix(1,2, "",k^2 ),k=1,n)+sum( matrix(1,2, "", k ),k=1,n)}}}



Combine the 1st and 3rd terms

{{{(n+1)sum( matrix(1,2, "", k ),k=1,n) - sum( matrix(1,2, "",k^2 ),k=1,n)}}} 

The first sum is the arithmetic series of the first n positive
integers,

which is given by the formula: {{{(n(n+1))/2}}}

so we have

{{{(n+1)((n(n+1))/2) - sum( matrix(1,2, "",k^2 ),k=1,n)}}} 

The remaining sum is the sum of the squares of the first n
positive integers.  This is also a well-known formula:

{{{(n(n+1)(2n+1))/6}}}

So now we have:

{{{(n+1)((n(n+1))/2) - (n(n+1)(2n+1))/6}}}

Rewrite the second term:

{{{(n+1)((n(n+1))/2) - (n(n+1)/2)((2n+1)/3)}}}

Factor out {{{(n(n+1))/2}}}

{{{((n(n+1))/2)((n+1)-(2n+1)/3)}}} 

Get LCD in the right parentheses:

{{{((n(n+1))/2)(3(n+1)/3-(2n+1)/3)}}}

{{{((n(n+1))/2)((3n+3)/3-(2n+1)/3)}}}

{{{((n(n+1))/2)((3n+3-(2n+1))/3))}}}

{{{((n(n+1))/2)((3n+3-2n-1)/3))}}}

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

{{{(n*(n+1)(n+2))/6}}}

Edwin</pre>