Waste less time on Facebook — follow Brilliant.
×

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
8 months, 1 week ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

There is also a "one-line" induction proof. Calvin Lin Staff · 8 months, 1 week ago

Log in to reply

There is a much simpler way to derive this formula.

Hint: Telescoping. Siddhartha Srivastava · 8 months, 1 week ago

Log in to reply

Comment deleted 8 months ago

Log in to reply

@Aditya Sky What is Pascal's rule ? ... yes as everyone says, the one line induction proof is much simpler Anirban Mandal · 8 months ago

Log in to reply

@Anirban Mandal 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 \(r \leq n\). Aditya Sky · 8 months ago

Log in to reply

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

\[\sum_{i=1}^N F_i=1+\sum_{i=2}^N F_{i-1}+F_{i-2}\\ \implies \sum_{i=1}^N F_i = 1+\sum_{i=1}^{N-1} F_i+\sum_{i=1}^{N-2} F_i = 1+\left(2\sum_{i=1}^N F_i\right)-(F_N+F_{N-1}+F_N)\\ \implies \sum_{i=1}^N F_i=(F_N+F_{N-1}+F_N)-1=(F_{N+1}+F_N)-1=F_{N+2}-1\] Prasun Biswas · 6 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...