Please disregard the below. I realized, upon further examination of the problem, that the function N has nothing to do with the behavior displayed. My apologies for the confusion.

I must confess that many of the mathematical elements of this problem fly over my head, since programming is more my forte, but I suspect your problem may have something to do with the fact the N is a predefined Mathematica function. See the below example, replacing N with d.

$Assumptions=d>=1&&d>=n>=0&&x\[Element]Reals&&{n,d}\[Element]Integers;
b-a
b-a//FullSimplify

I think this problem goes deeper than I originally thought, so i created a Notebook that should be easy to read and make the problem visible (attached bug.nb)

Here it can be seen that with the 2nd assumptions where "d" is an Integer, the expression

f1[n] - f2[n] //FullSimplify

is recognized to be 0 (which is true). The expression

a - Sum[ Binomial[d, n] f1[n] , {n,0,d} ] //FullSimplify

is not recognized to be 0, but

a - Sum[ Binomial[d, n] f2[n] , {n,0,d} ] //FullSimplify

is recognized to be 0.

Edit

Definitions:

a:= (Exp[x] + Exp[-x])^d
f1[n_] := Exp[(2 n - d) x]
f2[n_] := Exp[-x]^(d - n) Exp[x]^n

Assumptions 1 ( //FullSimplify correctly yields all zeros ):

d >= 1 && d >= n >= 0 && Element[x, Reals] && Element[n, Integers]

Assumptions 2 ( //FullSimplify does not yield a - Sum[ ... f1[n] ...] == 0 ):

d >= 1 && d >= n >= 0 && Element[x, Reals] && 
 Element[{n, d}, Integers]

Attachments: