let f(-1)=1, f(0)=0, f(1)=1, f(2)=1, f(3)=2, f(4)=3, f(5)=5, ..., f(n), ...

(a1)+(a2)+(a1+a2)+(a1+2a2)+(2a1+3a2)+(3a1+5a2)+(5a1+8a2)+...

nth term of previous summation is: (f(n-2)a1+f(n-1)a2)

sum of all n terms = (f(-1)+...+f(n-2))a1 + (f(0)+...+f(n-1))a2

but we have:

-- f0+f1 = f2 = f1+f2-f1 = f3-f1 = f3-1

-- f0+f1+f2 = f3-1+f2 = f4-1

-- by induction,   f0+f1+...+fn = f(n+2)-1

so:

== (f(-1)+...+f(n-2)) = 1 + f(0)+...+f(n-2) = 1 + f(n)-1 = f(n)

== (f(0)+...+f(n-1)) = f(n+1)-1

so:

sum of all n terms = (f(-1)+...+f(n-2))a1 + (f(0)+...+f(n-1))a2

sum of all n terms = (f(n))a1 + (f(n+1)-1)a2

sum of all n terms = f(n)a1 + f(n+1)a2 - a2