Proof about specific sum of Fibonacci numbers

Question

Let $F_k$ denote the $k$-th Fibonacci number. Find a formula for and prove by induction that your formula is correct for all $n > 0$. $$ (-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_n=\ ? $$ I have tried finding several formulas but all of them were wrong or where working only when n>1 please help me if you can

Answer 1

Using the famous closed form for the Fibonacci numbers and some power series, we can establish that your sum is the same as $$\frac{(-2)^{-n} (1+\sqrt{5})^{-n-1} (2 (1+\sqrt{5}))^{2n}+(3+\sqrt{5}) (-4)^n-(5+\sqrt{5}) (-2 (1+\sqrt{5}))^n)}{\sqrt{5}}$$ Although this will probably not satisfy you, it is irrefutably a closed form for this sum. It might be possible to transform it into a nicer one though...

Answer 2

Prove by induction that $$\sum_{i = 0}^{n}(-1)^{i}F_i = (-1)^nF_{n-1}-1$$

assuming that you have $F_{-1}=-1, F_0=0, F_1=1$

Answer 3

Let $A_n=\sum_{k=0}^{n}(-1)^{n-k} F_k$. Since: $$ \frac{x}{1-x-x^2}=\sum_{k\geq 0} F_{k}\,x^k,\tag{1} $$ we have: $$ A_n = [x^n]\left(\frac{1}{1+x}\cdot\frac{x}{1-x-x^2}\right),\tag{2} $$ but by partial fraction decomposition and $(1)$: $$ \forall n\geq 1,\qquad A_n = [x^n]\left(1-\frac{1}{1+x}+\frac{x^2}{1-x-x^2}\right) = F_{n-1}-(-1)^n\tag{3}$$ hence:

$$ \forall{n\geq 1},\qquad \sum_{k=0}^{n}(-1)^k F_k = (-1)^n F_{n-1}-1.\tag{4} $$

Answer 4

Assuming, $ F_{0} = 0 \ and\ F_{1} = 1 $.

We have two properties of Fibonacci numbers link

$ F_{1} + F_{3} + .... + F_{2*n-1} = F_{2*n} $

$ F_{2} + F_{4} + .... + F_{2*n} = F_{2*n+1} -1 $

$(-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_{2*n}= F_{2*n+1} - F_{2*n} - 1$

Hope this will help you.

Answer 5

Let $$ S_n=(-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_n. $$ If $F_0=F_1=1$ then $S_n=(-1)^n F_{n-1}+1$.