In my opinion, Maple simplifies the expression quite well. But what is your main goal? If you want a simple way to calculate this expression for specific parameters values  a, b, c, d  and a specific  n , then here is a simple procedure for this:

restart;
Sol:=proc(a,b,c,d,n0)
simplify(rsolve({f(n+1)-((a-1)*(b-1)+(c+d))*f(n)-c*d*f(n-1)=0, f(0)=a,f(1)=a*(b - 1)+c},f));
simplify(eval(%,n=n0));
end proc:	

Examples of use:

Sol(2,1,4,3,100); ``;
seq(Sol(2,1,4,3,k), k=1..20); ``;
Sol(2,1,4,3,n);

f := x -> 3*x + 2:
g := D(f):
g(3);

P:=proc(p,q,r)
local pq, pr, v, cv, T;
uses LinearAlgebra, plots, plottools;
pq:=arrow(p,q-p, width=0.1, color=red);
pr:=arrow(p,r-p, width=0.1, color=red);
v:=CrossProduct(convert(q-p,Vector),convert(r-p,Vector));
cv:=arrow(p,v, width=0.1, color=green);
T:=polygon([p,q,r], color=yellow);
display(pq,pr,cv,T, scaling=constrained, axes=normal, lightmodel=light1);
end proc:

Example of use:

P([3,3,2], [3,4,5], [4,3,4]);

If one solution is enough, then you can do so

restart;
FourSquares:=proc(n)
local a, b, c, d;
for a from 1 to floor(sqrt(n/4)) do
for b from a to floor(sqrt((n-a^2)/3)) do
for c from b to floor(sqrt((n-a^2-b^2)/2)) do
for d from c to floor(sqrt(n-a^2-b^2-c^2)) do
if n=a^2+b^2+c^2+d^2 then return [a,b,c,d] fi;
od: od: od: od:
end proc:

Example of use:

FourSquares(100000);
                                               [4, 8, 8, 316]

It is not recommended to use some variable and indexed variable with the same name in the same code. Therefore, I replaced, for example, t[n]  by  t||n  and so on. Some other corrections and additions were also made. The resulting approximate solution is quite accurate.

restart;
f:=t->y(t)-t^2+1;
eqn:=diff(y(t),t)=f(t):
ex:=dsolve({eqn,y(0)=0.5},y(t));
t||0:=0: w||0:=0.5: h:=0.2: ex||0:=0.5: e||0:=0:
Dy||0:=eval(f(t),[t=t||0,y(t)=w||0]);

for n from 1 to 10 do
t||n:=n*h; ex||n:=eval(rhs(ex),t=t||n);
w||n:=w||(n-1)+h*Dy||(n-1)+h^2/2!*(Dy||(n-1)-2*t||(n-1));
e||n:=abs(ex||n-w||n);
Dy||n:=eval(f(t),[t=t||n,y(t)=w||n]);
od:

printf(" i | t[i] |(Taylor)w[i] |(exact)y[i] |Error | \n ");
for i from 0 to 10 do
printf("%2.2f| %5.2f  | %5.6f| %5.6f  |  %5.6f | \n", i, t||i, w||i ,ex||i,e||i) ;
od;

A:=plot(rhs(ex), t=0..2, color=red):
B:=plot([seq([h*k,w||k], k=0..10)], style=point,color=blue):
plots:-display(A,B, legend=["Exact solution","Approximate solution"], view=0..6);

The final results as a table and a plot:

Unfortunately, it is not possible to obtain an explicit parameterization of the intersection curve, since finding x1,y1,z1 coordinates as functions of the parameter  theta  leads to the solution of an equation of a high degree. But we can get a numerical solution in the form of a procedure. This procedure  Curve  returns the coordinates of the point on the intersection curve for the specified values  theta, x, y, z :

Download Curve1.mw

Maple does not implement the solution of irrational inequalities with parameters. Here is a simpler example:

solve(sqrt(x-a)<x, x);

              Warning, solutions may have been lost

To calculate the values of derivatives of any order at a point, use  D  instead of  diff :

f:=x->x^2:
sum((D@@k)(f)(0), k=1..2);

As a workaround, use exact arifmetic:

answer := 5*exp(15/2*t);
response := subs(a = exp(1), 5*a^(15/2*t));
a := subs(t = Pi, answer);
b := subs(t = Pi, response);
simplify(a-b);

In order for Maple to solve this example step by step, we need to help it a little. To do this, first make the change  x=sin(t) . Maple shows a step-by-step solution, and in the end we do the inverse change and final simplification for a multiple argument:

A:=Int(x^2 * sqrt(1-x^2), x):
B:=IntegrationTools:-Change(A, x=sin(t));
Student:-Calculus1:-ShowSolution(B);
subs(t=arcsin(x), -(1/32)*sin(4*t)+(1/8)*t);
expand(%);

U:={1,2,3,4,5,6,7,8,9,10}:
A:={1,2,3,4}:
B:={4,5,6,7}:
C:={8,9,10}:
U minus C;
# Or in prefix notation
`minus`(U,C);

See the help page  Sets and Lists

If you already know the desired final result, then the easiest way is:

simplify(sin(4*Pi*w)/sin(2*Pi*w) - 2*cos(2*Pi*w));

Output :            0

The commands proposed in other answers are selected based on the already known result.
            

As an alternative you can use a  for-loop .

In the example below - the simplest way to plot the Pascal triangle:

N:=10:
for n from 0 to N do seq(binomial(n,k), k=0..n) od;

Acer's method does the same but the code is shorter a little:

N:=10:
seq(print(seq(binomial(n,k), k=0..n)), n=0..N);

plots:-intersectplot(x^2+y^2+z^2=1,x+y+z=0, x=-1..1, y=-1..1, z=-1..1);

This behavior seems like a bug. Here is a workaround:

sol_2:=subs(_C1=C[1], sol_1):
simplify(eval(ode, [sol_2, sol_2^2])) ;
                                 0=0