Letting $f_n$ be the Fibonacci numbers, show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer.

Just some homework help. Need to prove. Thank you in advance.

  • $\begingroup$ What is $f_n$? Fibonacci I assume.... $\endgroup$ – N. S. Nov 14 '11 at 21:32
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    $\begingroup$ Then it is not true ;) The problem should specify what $f_n$ is, and if not, it was probably defined earlier in another problem or textbook... $\endgroup$ – N. S. Nov 14 '11 at 21:36
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    $\begingroup$ Think about the problem the following way: pick any sequence which verifies this relation, pick any $n$ you want and then change ONLY $f_{2n}$... What happens to the equality? So, unless you know exactly what $f_n$ is for all $n$, you can't prove this is right... $\endgroup$ – N. S. Nov 14 '11 at 21:41
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    $\begingroup$ Yes, it is Fibonacci. Doesn't claim on the assinment, but does reference a set of problems from the book. My mistake, I apologize. $\endgroup$ – Chris Nov 14 '11 at 22:07
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    $\begingroup$ Yet another case where homework difficulties would better be solved by asking the instructor, rather than here! $\endgroup$ – GEdgar Nov 14 '11 at 22:16

Hint 1: Induction...

Hint 2: replace $f_n$ by their "closed" form, and see how you can calculate those sums.. It is easy...

Each hint leads to a different solution...

P.S. I assumed that $f_n$ is the Fibonacci sequence, I am pretty sure it is...

  • $\begingroup$ Thank you for the prompt answer, by the way. $\endgroup$ – Chris Nov 14 '11 at 22:56

I will give a less beautiful method, but it could be interesting and I think it's worth mentioning.

Let $F=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, with the property that $F^n=\begin{bmatrix}f_{n+1}&f_{n}\\f_{n}&f_{n-1}\end{bmatrix}$ for $n\in\mathbb Z$.

Consider $(-F)^0+(-F)^1+(-F)^2+\cdots+(-F)^{2n}$. This is $(-F-I_2)^{-1}((-F)^{2n+1}-(-F)^0)$.

A few calculations give $(-F-I_2)^{-1}=\begin{bmatrix}-1&1\\1&-2\end{bmatrix}$. Comparing the entries in the first row and second column on both sides, we see that the sum $f_0-f_1+f_2-\cdots+f_{2n}$ equals $$(-1)\cdot(-f_{2n+1})+1\cdot(-f_{2n}-1)=f_{2n-1}-1,$$ as desired.

Note that this method allows you to prove a closed form for such sums without knowing what the result should be.

  • $\begingroup$ I thought the matrix geometric series $I+A+A^2+\cdots=(I-A)^{-1}$ only converges when $\rho(A)<1$ but this matrix has the golden ratio as an eigenvalue. :/ $\endgroup$ – N8tron May 24 '18 at 13:26
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    $\begingroup$ @N8tron Here it's a finite sum, so no problem. $\endgroup$ – punctured dusk May 24 '18 at 13:28

In case you already now the formula for the sum of first $n$ Fibonacci numbers $$\sum_{j=0}^n F_j = F_{n+2} - 1$$ you can use it as follows:

$F_0-F_1+F_2-F_3+\dots-F_{2n-1}+F_{2n}=$ $F_0+(F_2-F_1)+(F_4-F_3)+\dots+(F_{2n}-F_{2n-1})=$ $0+F_0+F_2+\dots+F_{2n-2}=$ $0+(F_0+F_1)+\dots+(F_{2n-4}+F_{2n-3})=$ $\sum_{k=0}^{2n-3} F_n = F_{2n-1}-1.$

A proof of the above formula for the sum of the first $n$ Fibonacci numbers can be found e.g. at proofwiki.


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