What about 34234-2*17117 ?

BTW, F(a) = Kitonium's f(a),  but you think f is better.

@Kitonum 

This one is simpler and works for symbolic entries too:

F := a -> piecewise(is(a,odd),{(-a+1)/2,(-a-1)/2},{-round(a/2)});

Do you use a random generator for functions of two variables?

@guras 

@Jjjones98 

Then write S(n) as

Sum((1/k^0.1)*(sin(1/k)-1/k), k=1..n) +  Sum(1/k^(11/10), k=1..n);

The fist sum has a rapid convergence; for the second one use:

asympt(Sum(1/k^(11/10), k=1..n),n);
 

@Kitonum 

Actually the symbolic solution is probably useless (being too long) even without parametric and allsolutions.

You may be also interested in the related Apollonian circles
https://www.mapleprimes.com/questions/222071-Drawing-An-Apollonian-Gasket

Apollonius (of Perga) lived more than five centuries before Pappus.

Do you think that Maplesoft should invest a so much effort for typesetting?
Typesetting is a nice feature but I think that an excessive fine-tuning is too expensive.
I'd prefer these efforts be directed towards maths.
And anyway, Maple is not going to compete with LaTeX.
By the way, wouldn't it be more natural a LaTeX convention in the case of atomic variables?
It is not comfortable to use names like  `#msub(mi("x"),mn("1"))`  (in 1D notation)
(of course a macro() can be used here but even so...).

I also did not noticed the square. But Kitonum posted already a nice and complete solution.

P.S. It will be better in the future to post code instead of pictures.

@sajad 

@_Maxim_ 

I think you forgot index=1 in RootOf. Then it works as expected.
(It's only evalf which takes index=1 by default.)

@Mariusz Iwaniuk 

You cannot obtain this conclusion! The series could be divergent. for some c's.
As it is easy to check, the equation has at most 3 real roots,

@asa12 

Consider the row operation Eij of adding the row i to row j.
If  Eij . X = X  for each i,j (i<>j) then obviously X = O.

@asa12 

Your definition of "invariant matrix" is missing. For the standard definition, the only matrix is zero.

@_Maxim_ 

But if f is just absolutely integrable (in the Lebesgue sense) then the series could be divergent at any point! And anyway the side limits f(x0+0), f(x0-0) may not exist.