1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How the Fibonacci sequence is defined for you? $F_0 = 0, F_1=1$ or $F_0 = 1, F_1=1$ $\endgroup$ – sve Sep 16 '15 at 15:42
  • $\begingroup$ See this question and other questions linked there. $\endgroup$ – Martin Sleziak Oct 4 '15 at 7:00
1
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ thanks this was very helpful $\endgroup$ – Arman Piloyan Sep 16 '15 at 16:33
0
$\begingroup$

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...

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $S_n=(-1)^n F_{n-1}+1$.

$\endgroup$
  • $\begingroup$ How do you prove this? $\endgroup$ – vonbrand Sep 16 '15 at 16:34
  • $\begingroup$ By induction, for instance. $\endgroup$ – Aretino Sep 16 '15 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.