User:Bhpsngum

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

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

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:

And we have this result:

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 $$n \in \N^*$$)

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

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:

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

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