There surely is.
First, let's recall that the nth Fibonacci number equals
φ(n) = [φ^n - (-φ)^(-n)]/√5
where φ = (√5 + 1)/2 (Golden Ratio) and (-φ)^(-1) = (1-√5)/2. But to make this shorter, let me denote φ as A and (-φ)^(-1) as B.
Next, let's notice that a sum of Fibonacci numbers is a sum of powers of A and B:
[φ(n) + φ(m)]*√5 = A^n + A^m - B^n - B^m
Now what is enough to calc (in the {1,2,3}
example) is
A^1 + A^2 + A^3 + A^{1+2} + A^{1+3} + A^{2+3} + A^{1+2+3}.
But hey, there's a simpler expression for this:
(A^1 + 1)(A^2 + 1)(A^3 + 1) - 1
Now, it is time to get the whole result.
Let our set be {n1, n2, ..., nk}
. Then our sum will be equal to
Sum = 1/√5 * [(A^n1 + 1)(A^n2 + 1)...(A^nk + 1) - (B^n1 + 1)(B^n2 + 1)...(B^nk + 1)]
I think, mathematically, this is the "simplest" form of the answer as there's no relation between n_i. However, there could be some room for computative optimization of this expression. In fact, I'm not sure at all if this (using real numbers) will work faster than the "straightforward" summing, but the question was about avoiding subsets generation, so here's the answer.