Sum of Fibonacci

In this video, I calculate the sum of the first n Fibonacci numbers, using a neat telescoping sum-trick.

[eliyasne]

I used a telescoping series to but in a different way:
I just repeatedly used the definition of Fn

Fn+2=Fn+1 + Fn
=Fn+1 + Fn-1 + Fn-2
=Fn+1 + Fn-1 + Fn-3 + Fn-4
=Fn+1 + Fn-1 + Fn-3 + Fn-5 + Fn-6
And so on...
Fn+2 = Fn+1 + sum of (Fn-odds)
+ Fa
Where Fa is the tail of the sum.
Then i did the same thing to Fn+1 and substituted it into my equation
Fn+2 = sum of(Fn-evens)
+ sum of(Fn-odds) + Fa + Fb
= sum of(Fn) + junk
And finally i can find the "junk" by substituting n = 1
And i find "junk" = 1
Therefore
Sn = Fn+2 - 1

[VibingMath]

The formula at the end is so neat! Never think of using telescoping to simplify the terms before. Thank you Dr Peyam!

[ilprincipe]

I found an explicit formula, its kind of messy tho, i made a picture of it:

[ricardoguzman]

Very nice. I discovered a property of Fibonacci series that relates to powers of 2:

F(n-a)/F(n) x (√5 + 1)^a ≈ 2^a.
Check different values, it is an ultra cool formula.

[sirshaganguly]

I knew that method as Vn method...where you write the term Tn as the difference of two consecutive term say Vn-Vn-1 or Vn+1-Vn
and then add and cancel it really simplifies to find sum of some series
...thank you sir love your videos

[speechismyhammer]

Awesome video. I was not expecting the proof to be so simple. Math is so surprisingly elegant at times!

[jordantorres]

Clever! I was expecting some complex series, then you hit us with that simple formula.

[MrRyanroberson]

The metallic number Pn obeys the formula 1 + nPn = Pn^2. Pn = sqrt((n/2)^2 + 1) +n/2. This means: Pn^k = nPn^(k-1) + Pn^(k-2) = (n-1)Pn^(k-1) + (n+1)Pn^(k-2) + Pn^(k-3) = (n-1)Pn^k + nPn^(k-2) + nPn^(k-3)... down and down to (n+1)Pn + 1. therefore we add and subtract the irregular terms: in the front, we have this erroneous -Pn^k, and at the end we have the excess Pn+1 term. moving them, we have: (P^k + P^(k-1) - P - 1) = n * the sum from 1 to k-1 of P^i. divide both sides by n for the general formula: (Pn^k + Pn^(k-1) -Pn -1)/n = the sum of powers of Pn from 1 to k-1. Shift the index with m=k-1, and we can factor out a (Pn+1) to get (Pn+1)(Pn^m-1)/n = sum from 1 to m of Pn^i. Also note: For the special case of the golden ratio, Pn+1 = Pn^2, and n=1 so we get the surprisingly easy formula phi^m-1 = the sum from 1 to m of phi^i

[Uni-Coder]

Sum of fibonacci is almost fibonacci again. Integral of exp(x) is exp(x) again. So the fibonacci series is a discrete analog of exp(x) ?

[michaelz]

This is the first time I've ever heard the phrase "sum of the bunnies".

[chirayu_jain]

Thought of the general term . Then the sum of it would be a geometric series. But your method was neat 😃👍🏻❤️

[SR-kdwi]

CAN YOU PLEASE TEACH US GROUP THEORY FOR UNDERGRADS?

[leofisher]

you can prove this using the Binet formula and a finite geometric series and then applying some of the properties of the golden ratio to simplify

[plaustrarius]

Came to comment this exact statement @5:35 great video! thank you Dr. Peyam!

[OonHan]

This is like the powers of 2 where their sum to 2^n is 2^(n+1) - 1!!

[shahinjahanlu]

سال نو میلادی مبارک دکتر پیام عزیز. شاد و سلامت باشید

[theproofessayist]

So simple and so beautiful. Perhaps we should try an extension for the alternating series of Fibonacci numbers, product of Fibonacci numbers or something related to the Fibonacci numbers' cousin: the Lucas Numbers.

[orangeguy]

What about the infinite sum of 1/F_n? This series has eluded me for quite some time, have you ever come across a solution?

[spencersevilla]

I thought there is a shortcut to find the sum like arithmeic series and geometric series where you are going to find the sum of the sequence

[theeliminator]

2:40 idk why this made me laugh more than it should 😂