\[\color{red}{\large \sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}\].

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. \[\huge \color{blue}{\text{Proof :-}}\] By \(\color{green}{\text{Binet's Fibonacci Formula}}\), we know that : \(\color{red}{\large F_n\,=\,\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{2^{n} \cdot \sqrt5 \cdot }}\), 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 :- \[S \,=\, \frac{1}{\sqrt5} \left( \frac{(1+\sqrt{5})^{N+2}-(1-\sqrt{5})^{N+2}-2^{N+2}\sqrt{5}}{2^{N+2}} \right)\] \[S \,=\, \left(\frac{(1+\sqrt{5})^{N+2} - (1-\sqrt{5})^{N+2}}{2^{N+2} \cdot \sqrt{5}} \right) \,-\, 1 \] \[\large \boxed{\color{green}{\sum^{N}_{n=1}F_n \,=\, F_{N+2} -1}}\] Check this formula for this question.

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

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There is a much simpler way to derive this formula.

Hint: Telescoping.

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