Zrjydtj

I know this kind of question is frequent asked, yet each case has its own particularities. I will show my problem.

I define the following:

f[k_] := Binomial[n, k] (Sin[Φ]^2)^k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);

After that I want to perform the sum:

FullSimplify[Tan[Φ]^2*Sum[g[k]^2/f[n - k], k, 0, n],
Element[n, Integers] && n > 0 && Element[Φ, Reals] && 0 < Φ < Pi/2]

And I got the result:

4 n^2 Cos[Φ]^2 HypergeometricPFQ[-n,1-n/2-1/2 n Cos[2 Φ],1-n/2- 
1/2 n Cos[2 Φ],-n Cos[Φ]^2,-n Cos[Φ]^2,-Cot[Φ]^2] Sin[Φ]^(-2+2 n)

This as much as I can simplify the expression. However, according to book the result of the sum is just 4n

H[n_]=Tan[Φ]^2*Sum[g[k]^2/f[n - k], k, 0, n]]//FullSimplify

Table[H[i],i,1,10]//FullSimplify
4, 8, 12, 16, 20, 24, 28, 32, 36, 40

If I introduce the Assumptions in the following way at the beginning of the notebook:

$Assumptions = n ∈ Integers && n < 100 && 
 n > 0 && Φ ∈ Reals && 0 < Φ < Pi/2

The result of the simplification is yet worse.

The problem is that I have to perform some summations similar to this, but this time I don't have previous knowledge of the solution

Eliminating the trigonometric terms work in this case:

expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[-n,1-n/2-1/2 n Cos[2 Φ],1-n/2- 
 1/2 n Cos[2 Φ],-n Cos[Φ]^2,-n Cos[Φ]^2,-Cot[Φ]^2] Sin[Φ]^(-2+2 n);

FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]

4n

Clear["Global`*"];

$Assumptions =
 Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;

f[k_] := Binomial[n, k] (Sin[Φ]^2)^
 k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);

seq = Table[
 Tan[Φ]^2*Sum[g[k]^2/f[n - k], k, 0, n] // Simplify,
 n, 1, 10]

(* 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 *)

Using FindSequenceFunction,

H[n_] = FindSequenceFunction[seq, n]

(* 4 n *)

Verifying over wider range,

And @@ Table[
 H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], k, 0, n] // Simplify, n,
 1, 25]

(* True *)