# Interesting Fibonacci Identity

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

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. $\huge \color{#3D99F6}{\text{Proof :-}}$ By $\color{#20A900}{\text{Binet's Fibonacci Formula}}$, we know that : $\color{#D61F06}{\large F_n\,=\,\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{2^{n} \cdot \sqrt5 \cdot }}$, where $n\,\epsilon\,\mathbb{N}$. Say $S \,=\, \sum^{N}_{n=1}F_n$. So, $\large S \,=\, \sum^{N}_{n=1}\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n} }{2^{n} \cdot \sqrt5}$ $\implies S \,=\, \frac{1}{\sqrt5} \left( \frac{\frac{1+\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1+\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1+\sqrt5}{2}} \right) \,-\, \frac{1}{\sqrt5} \left( \frac{\frac{1-\sqrt5}{2}\,\cdot \, \left ( 1 \,-\, \left ( \frac{1-\sqrt5}{2} \right )^{N} \right )}{1 - \frac{1-\sqrt5}{2}} \right)$. This on simplification yields :- $S \,=\, \frac{1}{\sqrt5} \left( \frac{(1+\sqrt{5})^{N+2}-(1-\sqrt{5})^{N+2}-2^{N+2}\sqrt{5}}{2^{N+2}} \right)$ $S \,=\, \left(\frac{(1+\sqrt{5})^{N+2} - (1-\sqrt{5})^{N+2}}{2^{N+2} \cdot \sqrt{5}} \right) \,-\, 1$ $\large \boxed{\color{#20A900}{\sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}}$ Check this formula for this question.

5 years, 4 months ago

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

Staff - 5 years, 4 months ago

There is a much simpler way to derive this formula.

Hint: Telescoping.

- 5 years, 4 months ago