User:Bhpsngum

Derived a new formula from Binet's Formula

Simplification

Binet's formula can be expressed like this: (with n $\in \mathbb {N} ^{*}$)

$F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}}{\sqrt {5}}}$

Now, simplify the formula:

$F_{n}={\frac {\frac {(1+{\sqrt {5}})^{n}-(1-{\sqrt {5}})^{n}}{2^{n}}}{\sqrt {5}}}={\frac {(1+{\sqrt {5}})^{n}-(1-{\sqrt {5}})^{n}}{2^{n}{\sqrt {5}}}}$

let $A_{n}=(1+{\sqrt {5}})^{n}-(1-{\sqrt {5}})^{n}$, we have:

$F_{n}={\frac {A_{n}}{2^{n}{\sqrt {5}}}}\qquad (1)$

Rewrite $A_{n}$ in another form

Now, let's do some examples with n=1,2,3,4:

n	$(1+{\sqrt {5}})^{n}$	$(1-{\sqrt {5}})^{n}$	$A_{n}$
1	${\binom {1}{0}}1+{\binom {1}{1}}{\sqrt {5}}$	${\binom {1}{0}}1-{\binom {1}{1}}{\sqrt {5}}$	$2\times {\binom {1}{1}}{\sqrt {5}}$
2	${\binom {2}{0}}1+{\binom {2}{1}}{\sqrt {5}}+{\binom {2}{2}}({\sqrt {5}})^{2}$	${\binom {2}{0}}1-{\binom {2}{1}}{\sqrt {5}}+{\binom {2}{2}}({\sqrt {5}})^{2}$	$2\times {\binom {2}{1}}{\sqrt {5}}$
3	${\binom {3}{0}}1+{\binom {3}{1}}{\sqrt {5}}+{\binom {3}{2}}({\sqrt {5}})^{2}+{\binom {3}{3}}({\sqrt {5}})^{3}$	${\binom {3}{0}}1-{\binom {3}{1}}{\sqrt {5}}+{\binom {3}{2}}({\sqrt {5}})^{2}-{\binom {3}{3}}({\sqrt {5}})^{3}$	$2\times ({\binom {3}{1}}{\sqrt {5}}+{\binom {3}{3}}({\sqrt {5}})^{3})$
4	${\binom {4}{0}}1+{\binom {4}{1}}{\sqrt {5}}+{\binom {4}{2}}({\sqrt {5}})^{2}+{\binom {4}{3}}({\sqrt {5}})^{3}+{\binom {4}{4}}({\sqrt {5}})^{4}$	${\binom {4}{0}}1-{\binom {4}{1}}{\sqrt {5}}+{\binom {4}{2}}({\sqrt {5}})^{2}-{\binom {4}{3}}({\sqrt {5}})^{3}+{\binom {4}{4}}({\sqrt {5}})^{4}$	$2\times ({\binom {4}{1}}{\sqrt {5}}+{\binom {4}{3}}({\sqrt {5}})^{3})$
5	${\binom {5}{0}}1+{\binom {5}{1}}{\sqrt {5}}+{\binom {5}{2}}({\sqrt {5}})^{2}+{\binom {5}{3}}({\sqrt {5}})^{3}+{\binom {5}{4}}({\sqrt {5}})^{4}+{\binom {5}{5}}({\sqrt {5}})^{5}$	${\binom {5}{0}}1-{\binom {5}{1}}{\sqrt {5}}+{\binom {5}{2}}({\sqrt {5}})^{2}-{\binom {5}{3}}({\sqrt {5}})^{3}+{\binom {5}{4}}({\sqrt {5}})^{4}-{\binom {5}{5}}({\sqrt {5}})^{5}$	$2\times ({\binom {5}{1}}{\sqrt {5}}+{\binom {5}{3}}({\sqrt {5}})^{3}+{\binom {5}{5}}({\sqrt {5}})^{5})$
6	${\binom {6}{0}}1+{\binom {6}{1}}{\sqrt {5}}+{\binom {6}{2}}({\sqrt {5}})^{2}+{\binom {6}{3}}({\sqrt {5}})^{3}+{\binom {6}{4}}({\sqrt {5}})^{4}+{\binom {6}{5}}({\sqrt {5}})^{5}+{\binom {6}{6}}({\sqrt {5}})^{6}$	${\binom {6}{0}}1-{\binom {6}{1}}{\sqrt {5}}+{\binom {6}{2}}({\sqrt {5}})^{2}-{\binom {6}{3}}({\sqrt {5}})^{3}+{\binom {6}{4}}({\sqrt {5}})^{4}-{\binom {6}{5}}({\sqrt {5}})^{5}+{\binom {6}{6}}({\sqrt {5}})^{6}$	$2\times ({\binom {6}{1}}{\sqrt {5}}+{\binom {6}{3}}({\sqrt {5}})^{3}+{\binom {6}{5}}({\sqrt {5}})^{5})$

And we have this result:

n	$A_{n}$ simplified
1	$2{\sqrt {5}}\times {\binom {1}{1}}5^{0}$
2	$2{\sqrt {5}}\times {\binom {2}{1}}5^{0}$
3	$2{\sqrt {5}}\times ({\binom {3}{1}}5^{0}+{\binom {3}{3}}5^{1})$
4	$2{\sqrt {5}}\times ({\binom {4}{1}}5^{0}+{\binom {4}{3}}5^{1})$
5	$2{\sqrt {5}}\times ({\binom {5}{1}}5^{0}+{\binom {5}{3}}5^{1}+{\binom {5}{5}}5^{2})$
6	$2{\sqrt {5}}\times ({\binom {6}{1}}5^{0}+{\binom {6}{3}}5^{1}+{\binom {6}{5}}5^{2})$

To the examples, we have:

• n is odd: $A_{n}=2{\sqrt {5}}\times \sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{2k+1}}5^{k}$

• n is even: $A_{n}=2{\sqrt {5}}\times \sum _{k=0}^{\frac {n-2}{2}}{\binom {n}{2k+1}}5^{k}$

Hence:

$A_{n}=2{\sqrt {5}}\times \sum _{k=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }{\binom {n}{2k+1}}5^{k}\qquad (2)$

Final Steps and Conclusion

Substitute (2) to (1):

$F_{n}={\frac {2{\sqrt {5}}\times \sum _{k=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }{\binom {n}{2k+1}}5^{k}}{2^{n}{\sqrt {5}}}}$

Finally, we have a new formula derived from Binet's Formula:

$F_{n}={\frac {\sum _{k=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }{\binom {n}{2k+1}}5^{k}}{2^{n-1}}}$ (with $n\in \mathbb {N} ^{*}$)

Expand to negative numbers

The Fibonacci's sequence can also be extended to negative index n called "negafibonnaci" numbers below:

F−8

F−7

F−6

F−5

F−4

F−3

F−2

F−1

F0

F1

F2

F3

F4

F5

F6

F7

F8

−21

13

−8

5

−3

2

−1

1

0

1

1

2

3

5

8

13

21

We can see with n<0: $F_{n}<0$ when n is even and vice versa.

Hence, the $F_{n}$ formula can also express like this:

$F_{n}=({\frac {|2n+1|}{2n+1}})^{n+1}\times {\frac {\sum _{k=0}^{\left\lfloor {\frac {|n|-1}{2}}\right\rfloor }{\binom {|n|}{2k+1}}5^{k}}{2^{|n|-1}}}$

All formulae (with $n\in \mathbb {Z}$)

Closed-form

$F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}}{\sqrt {5}}}$

(Binet's Formula)

Summation formulae

$F_{n}=({\frac {|2n+1|}{2n+1}})^{n+1}\times {\frac {\sum _{k=0}^{\left\lfloor {\frac {|n|-1}{2}}\right\rfloor }{\binom {|n|}{2k+1}}5^{k}}{2^{|n|-1}}}$

(Derived from Binet's Formula)

$F_{n}=({\frac {|2n+1|}{2n+1}})^{n+1}\times \sum _{k=0}^{\left\lfloor {\frac {|n|-1}{2}}\right\rfloor }{\binom {|n|-k-1}{k}}$

(The sums of the "shallow" diagonals of Pascal's triangle)

with ${\binom {a}{b}}={\frac {a!}{b!(a-b)!}}$ and $\left\lfloor {\frac {|n|-1}{2}}\right\rfloor ={\frac {2|n|-3-(-1)^{n}}{4}}$