Fibbonacci Squares Proof
- Thread starter Goistein
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After some experimenting, I've found that the sum of the squares of consecutive fibbonacci numbers is equal to every other number.
So, F(n)^2+F(n+1)^2=F(2n+1)
Except I don't know the proof, so does anyone else know?
While I do not have the proof of your finding, I thought you might find the following of some interest.
Closer examination of the Fibonacci series will lead you to some surprisingly interesting and unique relationships between the Fibonacci numbers Fn = F(n-1) + F(n-2).
As n increases, the ratio of F(n+1)/Fn approaches the Golden Ratio, (1 + sqrt5)/2 = 1.618....
The sum of the squares of two adjacent Fibonacci numbers is equal to a higher Fibonacci number according to Fn^2 + F(n+1)^2 = F(2n+1). For instance, the 4thFn^2 + the 5thFn^2 = the F(2(4) + 1) = 9th Fn or 3^2 + 5^2 = 34, the 9th Fn.
The product of two alternating Fibonacci numbers minus the square of the one in between is equal to +/- one as expressed by F(n-1)F(N+1) - Fn^2 = (-1)^n. For instance, 8x21 - 13^2 = 168 - 169 = (-1)^7 = -1.
The sum of the cubes of two adjacent Fibonacci numbers minus the cube of the preceding one is equal to a higher Fibonacci number as expressed by Fn^3 + F(n+1)^3 = F3n. For instance, F4^3 + F5^3 - F3^3 = 3^3 + 5^3 - 2^3 = 27 + 125 - 8 = 144 = F12 = 144.
The sum of the squares of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to the product of Fn and F(n+1) as expressed by F1^2 + F2^2 + F3^2 + ......Fn^2 = FnF(n+1). For instance, 1^2 + 1^2 + 2^2 + 3^2 + 5^2 = 1 + 1 + 4 + 9 + 25 = 40 = 5x8.
The sum of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to a higher Fibonacci number as expressed by F1 + F2 + F3 + .......Fn = F(n+2) - 1. For instance, 1 + 1 + 2 + 3 + 5 + 8 = 20 = 21 - 1.
The sum of any 10 consecutive Fibonacci numbers is 11 times the 7th term of the 10 numbers.
For instance, the sum of the 4th through 13th numbers, 3,5,8,13,21,34,55,89,144,233, is 11x55 = 605.
The sum of any number of consecutive Fibonacci numbers is given by S[Fn1-->Fn2] = F(n2+2) - F(n1+1).
For instance, the sum of the 5th through 10th numbers, 5,8,13,21,34,55, is 144 - 8 = 136.
The sum of the first n even terms of Fibonacci numbers is given by F2 + F4 + F6 + F8 + ....F2n = F(2n+1) - 1.
For instance, the sum of the first 6 even terms is therefore F(2(6) + 1) - 1 = F(13) - 1 = 233 - 1 = 232.
The sum of the first n odd terms of Fibonacci numbers is given by F1 + F3 + F5 + ....F(2n-1) = F(2n)
For instance, the sum of the first 6 odd terms is therefore F(2n) = F(12) = 144.
Fn^2 + F(n+1)^2 = F(2n+1)
F(n+1)F(n-1) - Fn^2 = (-1)^n
Some other relationships of Fibonacci numbers are Fn^2 + F(n+1)^2 = F(2n+1) and F(n+1)F(n-1) - Fn^2 = (-1)^n.
The ratio of the 2nth Fibonacci number divided by the nth Fibonacci number is always an integer or F2n/Fn = K. For instance, F10/F5 = 55/5 = 11.
The sum of any 4n consecutive Fibonacci numbers is evenly divisible by F2n.
Example: The sum of 4(2) = 8 Fibonacci numbers is divisible by F2(2) = F4 = 3.
1+2+3+5+8+13+21+34 = 87/3 = 29.
2+3+5+8+13+21+34+55 = 141/3 = 47.
3+5+8+13+21+34+55+89 = 228/3 = 76
As any number may be represented by combinations of powers of 2, so may any number be represented by combinations of the Fibonacci numbers.
1 = 1, 2 = 2, or 1+1, 3 = 3 or 1+2, 4 = 3+1, 5 = 5 or 3+2, 6 = 5+1 or 3+2+1, 7 = 5+2, 8 = 5+3, 9=5+3+1, 10 = 5+3+2, 50 = 34+13+3, 100 = 89+8+3 or 55+34+8+3, and so on.
The greatest common divisor of any two Fibonacci numbers is, itself, a Fibonacci number. Even more surprising is the fact that the g.c.d.(Fa, Fb) = c = F[g.c.d.(a,b)]. What this means is that the g.c.d. of the "a"th and "b"th Fibonacci numbers is the "c"th Fibonacci number where c = the g.c.d. of a and b.
For any two consecutive Fibonacci numbers, a and b, a^2 - ab - b^2 = +1 or -1.
Given the four consecutive Fibonacci numbers a, b, c and d, (ad), 2(bc) and (cd - ab) form a Pythagorean Triple such that (ad)^2 + [2(bc)]^2 = (cd - ab)^2.
Given two positive integers "m" and "n" that are relatively prime, the Fibonacci numbers F(m) and F(n) are also relatively prime.
So, F(n)^2+F(n+1)^2=F(2n+1)
Except I don't know the proof, so does anyone else know?
While I do not have the proof of your finding, I thought you might find the following of some interest.
Closer examination of the Fibonacci series will lead you to some surprisingly interesting and unique relationships between the Fibonacci numbers Fn = F(n-1) + F(n-2).
As n increases, the ratio of F(n+1)/Fn approaches the Golden Ratio, (1 + sqrt5)/2 = 1.618....
The sum of the squares of two adjacent Fibonacci numbers is equal to a higher Fibonacci number according to Fn^2 + F(n+1)^2 = F(2n+1). For instance, the 4thFn^2 + the 5thFn^2 = the F(2(4) + 1) = 9th Fn or 3^2 + 5^2 = 34, the 9th Fn.
The product of two alternating Fibonacci numbers minus the square of the one in between is equal to +/- one as expressed by F(n-1)F(N+1) - Fn^2 = (-1)^n. For instance, 8x21 - 13^2 = 168 - 169 = (-1)^7 = -1.
The sum of the cubes of two adjacent Fibonacci numbers minus the cube of the preceding one is equal to a higher Fibonacci number as expressed by Fn^3 + F(n+1)^3 = F3n. For instance, F4^3 + F5^3 - F3^3 = 3^3 + 5^3 - 2^3 = 27 + 125 - 8 = 144 = F12 = 144.
The sum of the squares of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to the product of Fn and F(n+1) as expressed by F1^2 + F2^2 + F3^2 + ......Fn^2 = FnF(n+1). For instance, 1^2 + 1^2 + 2^2 + 3^2 + 5^2 = 1 + 1 + 4 + 9 + 25 = 40 = 5x8.
The sum of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to a higher Fibonacci number as expressed by F1 + F2 + F3 + .......Fn = F(n+2) - 1. For instance, 1 + 1 + 2 + 3 + 5 + 8 = 20 = 21 - 1.
The sum of any 10 consecutive Fibonacci numbers is 11 times the 7th term of the 10 numbers.
For instance, the sum of the 4th through 13th numbers, 3,5,8,13,21,34,55,89,144,233, is 11x55 = 605.
The sum of any number of consecutive Fibonacci numbers is given by S[Fn1-->Fn2] = F(n2+2) - F(n1+1).
For instance, the sum of the 5th through 10th numbers, 5,8,13,21,34,55, is 144 - 8 = 136.
The sum of the first n even terms of Fibonacci numbers is given by F2 + F4 + F6 + F8 + ....F2n = F(2n+1) - 1.
For instance, the sum of the first 6 even terms is therefore F(2(6) + 1) - 1 = F(13) - 1 = 233 - 1 = 232.
The sum of the first n odd terms of Fibonacci numbers is given by F1 + F3 + F5 + ....F(2n-1) = F(2n)
For instance, the sum of the first 6 odd terms is therefore F(2n) = F(12) = 144.
Fn^2 + F(n+1)^2 = F(2n+1)
F(n+1)F(n-1) - Fn^2 = (-1)^n
Some other relationships of Fibonacci numbers are Fn^2 + F(n+1)^2 = F(2n+1) and F(n+1)F(n-1) - Fn^2 = (-1)^n.
The ratio of the 2nth Fibonacci number divided by the nth Fibonacci number is always an integer or F2n/Fn = K. For instance, F10/F5 = 55/5 = 11.
The sum of any 4n consecutive Fibonacci numbers is evenly divisible by F2n.
Example: The sum of 4(2) = 8 Fibonacci numbers is divisible by F2(2) = F4 = 3.
1+2+3+5+8+13+21+34 = 87/3 = 29.
2+3+5+8+13+21+34+55 = 141/3 = 47.
3+5+8+13+21+34+55+89 = 228/3 = 76
As any number may be represented by combinations of powers of 2, so may any number be represented by combinations of the Fibonacci numbers.
1 = 1, 2 = 2, or 1+1, 3 = 3 or 1+2, 4 = 3+1, 5 = 5 or 3+2, 6 = 5+1 or 3+2+1, 7 = 5+2, 8 = 5+3, 9=5+3+1, 10 = 5+3+2, 50 = 34+13+3, 100 = 89+8+3 or 55+34+8+3, and so on.
The greatest common divisor of any two Fibonacci numbers is, itself, a Fibonacci number. Even more surprising is the fact that the g.c.d.(Fa, Fb) = c = F[g.c.d.(a,b)]. What this means is that the g.c.d. of the "a"th and "b"th Fibonacci numbers is the "c"th Fibonacci number where c = the g.c.d. of a and b.
For any two consecutive Fibonacci numbers, a and b, a^2 - ab - b^2 = +1 or -1.
Given the four consecutive Fibonacci numbers a, b, c and d, (ad), 2(bc) and (cd - ab) form a Pythagorean Triple such that (ad)^2 + [2(bc)]^2 = (cd - ab)^2.
Given two positive integers "m" and "n" that are relatively prime, the Fibonacci numbers F(m) and F(n) are also relatively prime.
Those last few are properties I've never even heard of! I really never would've guessed that the sequence was that special (I thought it just had that golden ratio property before). Where did you get these?
BTW, P.S. that was your 666th post
Oh, it's also my 100th
BTW, P.S. that was your 666th post
Oh, it's also my 100th