Interesting Fibonacci Identity

∑n=1NFn = FN+2−1\color{#D61F06}{\large \sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}.

I was trying to solve this problem. Whilst solving this question, I figured out a very interesting formula, which is as shown above.

I don't know whether this formula has already been discovered or not, but, anyways here is the proof. Proof :-\huge \color{#3D99F6}{\text{Proof :-}} By Binet’s Fibonacci Formula\color{#20A900}{\text{Binet's Fibonacci Formula}}, we know that : Fn = (1+5)n − (1−5)n2n⋅5⋅\color{#D61F06}{\large F_n\,=\,\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{2^{n} \cdot \sqrt5 \cdot }}, where n ϵ Nn\,\epsilon\,\mathbb{N}. Say S = ∑n=1NFnS \,=\, \sum^{N}_{n=1}F_n. So, S = ∑n=1N(1+5)n − (1−5)n2n⋅5\large S \,=\, \sum^{N}_{n=1}\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n} }{2^{n} \cdot \sqrt5}   ⟹  S = 15(1+52 ⋅ (1 − (1+52)N)1−1+52) − 15(1−52 ⋅ (1 − (1−52)N)1−1−52)\implies S \,=\, \frac{1}{\sqrt5} \left( \frac{\frac{1+\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1+\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1+\sqrt5}{2}} \right) \,-\, \frac{1}{\sqrt5} \left( \frac{\frac{1-\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1-\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1-\sqrt5}{2}} \right). This on simplification yields :- S = 15((1+5)N+2−(1−5)N+2−2N+252N+2)S \,=\, \frac{1}{\sqrt5} \left( \frac{(1+\sqrt{5})^{N+2}-(1-\sqrt{5})^{N+2}-2^{N+2}\sqrt{5}}{2^{N+2}} \right) S = ((1+5)N+2−(1−5)N+22N+2⋅5) − 1S \,=\, \left(\frac{(1+\sqrt{5})^{N+2} - (1-\sqrt{5})^{N+2}}{2^{N+2} \cdot \sqrt{5}} \right) \,-\, 1 ∑n=1NFn = FN+2−1\large \boxed{\color{#20A900}{\sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}} Check this formula for this question.

#NumberTheory

Note by Aditya Sky
5 years, 2 months ago

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Comments

There is also a "one-line" induction proof.

Calvin Lin Staff - 5 years, 2 months ago

There is a much simpler way to derive this formula.

Hint: Telescoping.

Siddhartha Srivastava - 5 years, 2 months ago
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