@Kitonum   We can solve the problem numerically by reducing it to the calculation of the sum a series:

f:=n->sin(sqrt(n^2+n)*Pi)^2;

evalf[100](sum(f(n+1) - f(n), n=1..infinity)+f(1));

@Carl Love  I think your solution closer to hand than to automatic. Why we have to guess to use asympt(sqrt(n^2+n)*Pi, n, 1) ?  We wish the developers to solve this problem in a future release of Maple.

BTW in Mathematica the problem already solved:

Limit[Sin[Sqrt[n^2 + n]*Pi]^2, n -> Infinity, Assumptions -> Element[n, Integers]]

                                                                    1

@tomleslie  Thanks. And how about this example for  n::posint  in Maple:

limit(sin(sqrt(n^2+n)*Pi)^2, n=infinity);

@Markiyan Hirnyk  Why did you write the option "solutions=16" ? Without this option, DirectSearch finds some the other number of solutions. I think that aser's and Preben's ways (using only Maple tools) are more  justified.

@Carl Love   Of course, you're right! Still, the list is somewhat broader. As it is written in the help, for example, the list includes the values of functions with rational arguments and algebraic numbers of degree <=6.

Examples:

Digits:=20:

evalf(ln(101));

identify(%);

convert(%, ln);

evalf(sqrt(3+5^(1/3)));

identify(%);

evalf(sin(1/3));

identify(%);

Markiyan Hirnyk  Please provide a link to the source of this example. It is very similar to the challenges that are offered on student competitions.

@Markiyan Hirnyk  In fact squaring - this is a very natural idea, which is often useful. Here is another solution without complex numbers and conversions:

expr:=expand(sin(-(1/6)*Pi+(1/2)*arccos(1/3)));

z:=1/2*arccos(1/3):  # Obviously  cos(z)>0  and  sin(z)>0

eval(expr, [cos(z)=sqrt(combine(cos(z)^2)), sin(z)=sqrt(combine(sin(z)^2))]);

@Markiyan Hirnyk   The ratio of axes of the ellipse is equal to

sqrt(160*(1/4))/sqrt(160/9);

                        3/2

@Axel Vogt  The basic ideas for writing this procedure, I found here 

@Markiyan Hirnyk   Оbviously  999999999999999999990 = 10^21-10  rather than  10^20+10^20-10

With Maple:

is(999999999999999999990 = 10^21-10);

is(20*9^20 > 10^21-10);

                          true

                          false

@Carl Love  Thanks. Yes you are right. I did not notice that   _Z6  and  _Z7  can be nonpositive. Therefore, solutions will be more than 106141495. I will think how to fix this situation.

@ramakrishnan   The latter calculation  5!/(5-1-4)! * 4!/(1 * 4)! * 1/5^4 *1/2 ^2  is wrong.

Must be  

5!/(5-0-2)! * 4!/(1 * 2)! * 1/5^4 *1/2^2;

@Markiyan Hirnyk   All right. Make restart  before substitution:

restart;

eval(30*a+75*b+110*c+85*d+255*e+160*f+15*g+12*h+120*i=8000, [a = 2, b = 4, c = 6, d = 5, e = 16, f = 9, g = 1, h = 5, i = 8]);

                                                                    8000 = 8000

@Carl Love 

M:=(N::nonnegint, i::nonnegint, alpha) -> Matrix(N+1, (k,j)->`if`(j=k+i,1/GAMMA((k-1)*alpha+1), 0)):

@Rouben Rostamian   I did in the classic interface Maple 12.