1
$\begingroup$

each even term in the Fibonacci sequence has a position index which is a multiple of 3, therefore the even term is $F_{3n}$:

$F_{3 (0)} = 0$

$F_{3 (1)} = 2$

$F_{3 (2)} = 8$

$F_{3 (3)} = 34$

$F_{3 (4)} = 144$

...

How can I prove and reconstruct the formula which states that the sum of the first $n$ even terms (2, 8, 34, ...) of the Fibonacci sequence is given by $(F_{3n} + F_{3n + 3} - 2)/4$ not using induction?

For example, given:

$U_{n} = F_{3n}$

$S_{n} = U_{0} + U_{1} + U_{2} + U_{3} + ... + U_{n}$

How to derive the formula $S_n = (U_{n} + U_{n + 1} - 2) / 4 = \frac{(F_{3n} + F_{3n + 3} - 2)}{4}$?

$\endgroup$
  • $\begingroup$ Have you tried finding a recursive formula for this sequence similar to the one defining the Fibonacci numbers? $\endgroup$ – Nate May 4 '15 at 18:55
  • $\begingroup$ @Nate Should we say that using that recurrence relation is also using induction? I think the only possible way to avoid induction is to use the close form formula. $\endgroup$ – Salomo May 4 '15 at 18:57
3
$\begingroup$

It is relatively easy to show that , for $ n \geqslant 1 $, $4F_{3n} = F_{3(n+1)}-F_{3(n-1)}$(Hint: Use the definition of the Fibonacci numbers). Using this identity, we can now show that $$\sum_{k=1}^n F_{3k} = \frac{(F_{3n} + F_{3n + 3} - 2)}{4} $$

by summing up $4F_{3k}$, where k goes from $1$ to $n$.

$$\begin{align} 4 \sum_{k=1}^n F_{3k} &= \sum_{k=1}^n 4F_{3k}= \sum_{k=1}^n F_{3(k+1)}-F_{3(k-1)} \\ &= \sum_{k=1}^n F_{3(k+1)} - \sum_{k=1}^n F_{3(k-1)} \\&= \sum_{k=2}^{n+1} F_{3k} - \sum_{k=0}^{n-1} F_{3k} \\ &= F_{3n} + F_{3n + 3} + \sum_{k=2}^{n-1} F_{3k} - \sum_{k=2}^{n-1} F_{3k} -2 \\ &= F_{3n} + F_{3n + 3} - 2 \end{align}$$

Dividing everything by $4$ , we reach the desired equality.

$\endgroup$
  • $\begingroup$ Sorry for the late response, thanks for the passages! $\endgroup$ – user3019105 May 7 '15 at 8:23
1
$\begingroup$

What is being asked is to find the sum \begin{align} \sum_{r=0}^{n} F_{3r}. \end{align} This is obtained in the following way. \begin{align} \sqrt{5} \, S_{n} &= \sum_{r=0}^{n} \left( \alpha^{3r} - \beta^{3r} \right) \\ &= \frac{1 - \alpha^{3n+3}}{1-\alpha^{3}} - \frac{1-\beta^{3n+3}}{1-\beta^{3}} \\ &= - \frac{1}{2\alpha} + \frac{\alpha^{3n+2}}{2} + \frac{1}{2 \beta} - \frac{\beta^{3n+2}}{2} \\ &= \frac{\sqrt{5}}{2} \left( F_{3n+2} -1 \right). \end{align} From this it can be stated that \begin{align} \sum_{r=0}^{n} F_{3r} = \frac{F_{3n+2} - 1}{2}. \end{align}


The formulas used in this evaluation are: \begin{align} F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta} \end{align} which is Binet's formula and \begin{align} \sum_{r=0}^{n} x^{r} = \frac{1 - x^{n+1}}{1-x}. \end{align} Other components used: \begin{align} 1 - \alpha^{3} &= 1 - \alpha(1 + \alpha) = 1 - \alpha - (1 + \alpha) = - 2 \alpha \\ 1 - \beta^{3} &= - 2 \beta \\ \alpha \beta &= -1 \end{align}

$\endgroup$
  • $\begingroup$ I didn't fully understand all the passages, could you give me some links or explain more thorough the passages you took? I know that $(α^{3r}−β^{3})$ comes from the Binet's formula, where $α = (1 + \sqrt{5})/2$ and $β = -1/α$. In the following passages you are using some properties of the geometric series, aren't you? $\endgroup$ – user3019105 May 4 '15 at 19:29
  • $\begingroup$ @user3019105 The components used have been added. Note that $\alpha^{2} = 1 + \alpha$ and the same applies to $\beta$. $\endgroup$ – Leucippus May 4 '15 at 19:54
  • $\begingroup$ Thanks for the elucidation, your formula is really really nice! $\endgroup$ – user3019105 May 7 '15 at 21:05

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.