Question

If anyone can fully explain this to me I will greatly appreciate it

Answer 1

explain what lol

Answer 2

Prove that \[f_{2n+1}=f^{2}_{n+1}+f^{2}_{n}\]
Where n is a positive integer

Answer 3

Do you have the function f?

Answer 4

its the fibonacci number and i have to prove by induction

Answer 5

The Fibonacci sequence is defined by:
\[f_{1}=1, f_{2}=1\]
\[f_{n}=f_{n-1}+f_{n-2}, n \ge3\]

Answer 6

i bet we can do this. give me a second with some scrap paper

Answer 7

Thank you, I just cant seem to get a handle on this!

Answer 8

I'm sure i had to work this once before...but it's been quite some time. satellite, i'm sure will be much faster at it than i. :)

Answer 9

before i write it out idea is probably this. you get to assume (by induction) that
\[f_2n=f_n^2+f_{n-1}^2\]

Answer 10

and you also know that
\[f_{n+1}^2=(f_n+f_{n-1})^2\]

Answer 11

so i think those two facts combined with some algebra, might do it.

Answer 12

not saying that it will be fast either! my algebra is slow sometimes, but i think if you expand and combine terms correctly we can get what we want. now i will try it

Answer 13

How did you know \[f^{2}_{n+1}=(f_{n}+f_{n-1})^2\]

Answer 14

is that an identity?

Answer 15

that is because
\[f_{n+1}=f_n+f_{n-1}\]

Answer 16

gotcha

Answer 17

satellite73 one more question :)

Answer 18

ok so much for my hubris. this think is called lucas identity and i really have no idea how to prove it, but i think it is not trivial. do you have any hints?

Answer 19

i am going to find a proof of this, i don't care how long it takes

Answer 20

so determined :)

Answer 21

ok it seems that this is generally known and it also seems that i am not only not smart enough to come up with it on my own i am also not smart enough to find it on line. i did find this
http://answers.yahoo.com/question/index?qid=20070912044816AA2bzKm
which i will attempt to decipher.
hello joemath. got a reference or know this? i found some combinatoric proof that depends on tiling but that cannot be necessary

Answer 22

maybe you can use the closed formula for the Fibonacci sequence instead? The nth Fibonacci number is:
\[F(n) = \frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n}-(\frac{1-\sqrt{5}}{2})^{n})\]

Answer 23

unfortunately im reading for an american lit class so i cant sit down and work out details >.< (as much as i want to! this seems like an interesting problem!)

Answer 24

but im thinking if you work with F(2n+1), you might be able to factor/change it to (F(n+1))^2+F(n)^2 (or the other way around)

Answer 25

if this question is still open after my classes i'll will gladly start working on it :)

Answer 26

got it
@joemath it does not use closed form because he was asked to do it by induction. gimmick is to consider
\[f_{n+k}\] and induct on k not n.
i got it not by being smart but by going over the proof slowly.

Answer 27

nice :) sry i didnt read that part about doing it by induction >.<