User:Graeme McRae/Sum of 2m Consecutive Fibonacci Numbers

The sum of an even number of consecutive Fibonacci numbers is the product of a Lucas number and a Fibonacci number.

Lemmas

From Miklos Kristof, Mar 19 2007, a comment in A000045 : (Start)

Let L(n)=A000032=Lucas numbers. Then:

For a>=b and odd b, F(a+b)+F(a-b)=L(a)*F(b).

For a>=b and even b, F(a+b)+F(a-b)=F(a)*L(b).

For a>=b and odd b, F(a+b)-F(a-b)=F(a)*L(b).

For a>=b and even b, F(a+b)-F(a-b)=L(a)*F(b).

F(n+m)+(-1)^m*F(n-m)=F(n)*L(m);

F(n+m)-(-1)^m*F(n-m)=L(n)*F(m);

F(n+m+k)+(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k))=F(n)*L(m)*L(k);

F(n+m+k)-(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k))=L(n)*L(m)*F(k);

F(n+m+k)+(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k))=L(n)*F(m)*L(k);

F(n+m+k)-(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k))=5*F(n)*F(m)*F(k). (End)

Sum of 2m Consecutive Fibonacci Numbers

Proof that for n>m, the sum of the 2m consecutive Fibonacci numbers F(n-m-1) thru F(n+m-2) is F(n)*L(m) if m is odd, and L(n)*F(m) if m is even.

(1) It's true for m=1 -- F(n-2)+F(n-1)=F(n)=F(n)*L(1)

(2) It's true for m=2 -- F(n-3)+F(n-2)+F(n-1)+F(n)=L(n)=L(n)*F(2)

(3) For odd m, suppose F(n-m-1)+F(n-m)+...+F(n+m-2)=F(n)*L(m)

(4) Then from Kristof 2007, F(n-m-3)+F(n+m-1)=F(n-2)*L(m+1) with a=n-2 and b=m+1

(5) Similarly, F(n-m-2)+F(n+m)=F(n-1)*L(m+1) with a=n-1 and b=m+1

(6) Adding (3),(4),(5), F(n-m-3)+...+F(n+m)=F(n)*L(m)+F(n-2)*L(m+1)+F(n-1)*L(m+1)

(7) The RHS of (6) becomes F(n)*L(m)+F(n)*L(m+1) = F(n)*L(m+2), proving this case.

(8) For even m, suppose F(n-m-1)+F(n-m)+...+F(n+m-2)=L(n)*F(m)

(9) Then from Kristof 2007, F(n-m-3)+F(n+m-1)=L(n-2)*F(m+1) with a=n-2 and b=m+1

(10) Similarly, F(n-m-2)+F(n+m)=L(n-1)*F(m+1) with a=n-1 and b=m+1

(11) Adding (8),(9),(10), F(n-m-3)+...+F(n+m)=L(n)*F(m)+L(n-2)*F(m+1)+L(n-1)*F(m+1)

(12) The RHS of (11) becomes L(n)*F(m)+L(n)*F(m+1) = L(n)*F(m+2), proving this case.