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# Interesting Fibonacci Identity

$\color{red}{\large \sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}$.

I was trying to solve this problem. Whilst solving this question, I figured out a very interesting formula, which is as shown above.

I don't know whether this formula has already been discovered or not, but, anyways here is the proof. $\huge \color{blue}{\text{Proof :-}}$ By $$\color{green}{\text{Binet's Fibonacci Formula}}$$, we know that : $$\color{red}{\large F_n\,=\,\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{2^{n} \cdot \sqrt5 \cdot }}$$, where $$n\,\epsilon\,\mathbb{N}$$. Say $$S \,=\, \sum^{N}_{n=1}F_n$$. So, $$\large S \,=\, \sum^{N}_{n=1}\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n} }{2^{n} \cdot \sqrt5}$$ $\implies S \,=\, \frac{1}{\sqrt5} \left( \frac{\frac{1+\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1+\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1+\sqrt5}{2}} \right) \,-\, \frac{1}{\sqrt5} \left( \frac{\frac{1-\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1-\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1-\sqrt5}{2}} \right)$. This on simplification yields :- $S \,=\, \frac{1}{\sqrt5} \left( \frac{(1+\sqrt{5})^{N+2}-(1-\sqrt{5})^{N+2}-2^{N+2}\sqrt{5}}{2^{N+2}} \right)$ $S \,=\, \left(\frac{(1+\sqrt{5})^{N+2} - (1-\sqrt{5})^{N+2}}{2^{N+2} \cdot \sqrt{5}} \right) \,-\, 1$ $\large \boxed{\color{green}{\sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}}$ Check this formula for this question.

Note by Aditya Sky
1 year, 7 months ago

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## Comments

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There is also a "one-line" induction proof.

Staff - 1 year, 7 months ago

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There is a much simpler way to derive this formula.

Hint: Telescoping.

- 1 year, 7 months ago

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Comment deleted Apr 07, 2016

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What is Pascal's rule ? ... yes as everyone says, the one line induction proof is much simpler

- 1 year, 7 months ago

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I didn't know that such an identity existed. As I said, I just got this while solving a problem. And, sorry, this identity has nothing to do with Pascal's Rule. I wanted to refer to the recursive relation that describes Fibonacci's Sequence. Anyways, Pascal's Rule :- $$\color{green}{\binom{n}{r}\,+\,\binom{n}{r-1}\,=\, \binom{n+1}{r}}$$, where $$r$$ and $$n$$ are non-negative integers such that $$r \leq n$$.

- 1 year, 7 months ago

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As Siddharth mentions, the telescoping method should be a simpler direct proof method.

$\sum_{i=1}^N F_i=1+\sum_{i=2}^N F_{i-1}+F_{i-2}\\ \implies \sum_{i=1}^N F_i = 1+\sum_{i=1}^{N-1} F_i+\sum_{i=1}^{N-2} F_i = 1+\left(2\sum_{i=1}^N F_i\right)-(F_N+F_{N-1}+F_N)\\ \implies \sum_{i=1}^N F_i=(F_N+F_{N-1}+F_N)-1=(F_{N+1}+F_N)-1=F_{N+2}-1$

- 1 year, 5 months ago

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