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
3 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Calvin Lin Staff - 3 years ago

Log in to reply

There is a much simpler way to derive this formula.

Hint: Telescoping.

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...