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This is from problem 20

Is there a simplified expression for sum of Fibonacci numbers: $$F(a)+F(a+4)+F(a+8)+\ldots + F(a+4m)\;\;?$$ https://resources.thiel.edu/mathproject/atps/PDF/Chapt02.PDF It is problem 20. I am thinking that maybe I could cancel out certain terms, F(a)+F(a+8)=7F(a+4). But I don't know how to do nex

Xia
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    You should put your thoughts in the body of the question instead of the comments. – saulspatz Jul 31 '18 at 23:24
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    Maybe Binet's formula will give you a proper answer? – Jakobian Jul 31 '18 at 23:25
  • Since $F_{a+4k}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^a\left(\frac{1+\sqrt{5}}{2}\right)^{4k}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^a\left(\frac{1-\sqrt{5}}{2}\right)^{4k}$, then you can add for $k=0$ to $k=m$ to get $\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^a\cdot\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{4(m+1)}-1}{\left(\frac{1+\sqrt{5}}{2}\right)^{4}-1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^a\cdot\frac{\left(\frac{1-\sqrt{5}}{2}\right)^{4(m+1)}-1}{\left(\frac{1-\sqrt{5}}{2}\right)^{4}-1}$ –  Jul 31 '18 at 23:53
  • What is $F$? Linking is not enough, insert from the remote content so many which makes your question comprehensible even if the remote link disappears. It is a rule here. – peterh Jul 31 '18 at 23:59

2 Answers2

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Here is an answer using my favorite Fibonacci technique.

The matrix formulation for the Fibonacci sequence is well worth knowing and easily proved: $$ \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$

Let $ A=\begin{pmatrix}1&1\\1&0\end{pmatrix} $ and let's compute $$ S = A^a + A^{a+4} + \cdots + A^{a+4m} = A^a(I+A^4+\cdots+A^{4m}) = A^a(A^{4m+4}-I)(A^4-I)^{-1} $$ The sum we seek is the $(1,2)$ entry in $S$: $$ S= \begin{pmatrix}F_{a+1}&F_{a}\\F_{a}&F_{a-1}\end{pmatrix} \begin{pmatrix}F_{4m+5}-1&F_{4m+4}\\F_{4m+4}&F_{4m+3}-1\end{pmatrix} \begin{pmatrix}4&3\\3&1\end{pmatrix}^{-1} $$

lhf
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Render

$F_n=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$

Then your series is the difference between two geometric series and can be summed accordingly.

Oscar Lanzi
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