vv

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If you don't want to read at least a tutorial, the shortest path would be to post here a simplified version of your problem.
Probably someone will upload a worksheet with the solution, and you will be able to adapt it for similar tasks.

@Carl Love 

This order is not total (linear) and has nothing to do with "lexicographic". Where is it used? How is x[2] computed using this order?

x[n] seems to be the least positive interger not in { x[1],...,x[n-1] } which satisfies ...

Do you mean the lexicographic order in N? But then 123 < 2 < 3, so why is x[2] = 3?

It is not only old code, but you have also replaced the old keyword array with the new one Array in a few places. Revert these changes.

@mmcdara The first solution is very elegant; of course I like it. But it was probably provided by the author/committee or maybe by a very smart student (who deserves congratulations & a special prize, if he/she exists).

I don't know what experience you have in this field, but I was involved in these competitions
as a direct participant (a long time ago) and then as a team leader for my university students
(I have even proposed problems for the IMC contest).

Now I don't have such preoccupations but I still enjoy to occasionally watch these competitions
and find solutions (with or without Maple).

 

I think it's better to simplify first by hand.
Note first that expr = 0 for r>n (both terms are 0).
Then express everything w.r.t.
y = q^r;
x = q^n;
Y = product(y - q^i, i = 1 .. r - 1);
X = product(x - q^i, i = 1 .. r - 1);

and finally, simplify (or factor) the result (==> rational function in q,x,y,X,Y).

Maple simplifies telescoping sums:

ss:=sum(a[i],i=1..r) - sum(a[i+1],i=1..r):
combine(ss);

                        -a[r + 1] + a[1]
but not telescopic products.
pp:=product(a[i],i=1..r) / product(a[i+1], i=1..r);
 
Actually, combine works only if we rewrite
pp:=product(a[i],i=1..r) * product(a[i+1]^(-1), i=1..r);
 
to obtain
      product(a[i]/a[i+1],i = 1 .. r)
but it is not simplified further to a[1] / a[r+1].
The only solution seems to be a simple custom procedure simp_telescopic.
 

@nm You are confusing the inverse of a function with multiplicative inverse.

@nm 

Of course I would not classify y = g(y') as d'Alembert, but it matches the pattern y = x*f(y') + g(y') for f = 0
and the solution is valid for f=0 too. Anyway, the classification is mainly didactic and Maple does non necessarily use the classical methods. So, your problem is more or less artificial.

@acer Ok, so, a bug in IntegrationTools:-Expand (2015).

@mthkvv 

modp1 is builtin, so its code is not accessible. You will need to be employed by Maplesoft :-)

@gawati2611 

https://www.maplesoft.com/applications/view.aspx?SID=33406

(it is the first result with any search engine)

@gawati2611

You must download (from Maple Application Center) and install the package. 

I think you are supposed to do the computations by hand (it's very easy) and then check them with Maple.
If you don't, you will never know linear algebra!

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