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

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

There is a much simpler way to derive this formula.

Hint: Telescoping.

Log in to reply

Comment deleted Apr 07, 2016

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply