But I'm Not Listing All Of Them!

Let f n f_n denote the n th n^\text{th} Fibonacci number .
Given that f 22 = 17711 f_{22} = 17711 , calculate i = 1 20 f i \displaystyle \sum_{i = 1}^{20}f_i .


The answer is 17710.

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2 solutions

Rishabh Jain
Apr 5, 2016

Relevant wiki: Fibonacci Sequence

Let S n = r = 1 n f n \mathbf{S_n}=\displaystyle\sum_{r=1}^nf_n .

S n = 1 + 1 + 2 + + f n S n = 1 + 1 + + f n 1 + f n \begin{aligned}\mathbf{S_n}=&1+1+2+\cdots +\color{#D61F06}{f_n}\\\mathbf{S_n}=&~~~~~~~1+1+\cdots+\color{#D61F06}{f_{n-1}}+\color{#3D99F6}{f_n}\end{aligned}

Adding and using f n + f n 1 = f n + 1 \color{#D61F06}{f_{n}+f_{n-1}}=\color{#3D99F6}{f_{n+1}} and then using f n + 1 + f n = f n + 2 \color{#3D99F6}{f_{n+1}+f_{n}}=f_{n+2}

2 S n = 1 + 2 + 3 + + f n S n 1 + f n + 2 2\mathbf{S_n}=\underbrace{1+2+3+\cdots+f_n}_{\large \mathbf{S_n}~-1}+ f_{n+2}

S n = f n + 2 1 \large \implies\mathbf{S_n}=f_{n+2}-1

Put n = 20 n=20 to get: S 20 = 17710 \Large \mathbf{S_{20}}=\Large\boxed{17710}

Otto Bretscher
Apr 5, 2016

Let's prove k = 1 n f k = f n + 2 1 \sum_{k=1}^{n}f_k=f_{n+2}-1 by induction: k = 1 n + 1 f k = f n + 1 + k = 1 n f k = f n + 1 + f n + 2 1 \sum_{k=1}^{n+1}f_k=f_{n+1}+\sum_{k=1}^{n}f_k=f_{n+1}+f_{n+2}-1 = f n + 3 1 =f_{n+3}-1 . Thus the answer is f 22 1 = 17710 f_{22}-1=\boxed{17710}

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