Question 1207587
<pre>
I finally figured out that the above, which was gobbledygook to me, 
means this:
</pre>
The sum
{{{6(1*1+2*2+""*""*""*""+n*n)}}}
is equal to a polynomial f(n) for all {{{n>=1}}}.
Write f(n) as a polynomial with terms in descending order of n.
<pre>
It is well-known that

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

so

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

So

{{{"f(n)"}}}{{{""=""}}}{{{2n^3+3n^2+n}}}

Edwin</pre>