Sum of nth Fibonacci Number over nth Power of 2/Proof 2
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Theorem
- $\ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n} = 2$
where $F_n$ is the $n$th Fibonacci number.
Proof
From the Euler-Binet Formula, we have:
- $F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5}$
Therefore:
| \(\ds \sum_{n \mathop = 0}^\infty \frac{F_n} {2^n}\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5 \times 2^n}\) | Euler-Binet Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {\dfrac {1 + \sqrt 5} 2}^n - \paren {1 - \paren {\dfrac {1 + \sqrt 5} 2 } }^n} {\sqrt 5 \times 2^n}\) | Definition of Golden Mean | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {\dfrac {1 + \sqrt 5} 2}^n - \paren {\dfrac {1 - \sqrt 5} 2 }^n} {\sqrt 5 \times 2^n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt 5} \sum_{n \mathop = 0}^\infty \paren {\dfrac {1 + \sqrt 5} 4}^n - \dfrac 1 {\sqrt 5} \sum_{n \mathop = 0}^\infty \paren {\dfrac {1 - \sqrt 5} 4}^n\) | Product of Powers and $2^2 = 4$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \dfrac {1 + \sqrt 5 } 4} } - \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \dfrac {1 - \sqrt 5} 4} }\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \dfrac {1 + \sqrt 5} 4} } \times \dfrac 4 4 - \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \dfrac {1 - \sqrt 5} 4} } \times \dfrac 4 4\) | multiplying top and bottom by $4$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 4 {\sqrt 5} \paren {\dfrac 1 {3 - \sqrt 5} } - \dfrac 4 {\sqrt 5} \paren {\dfrac 1 {3 + \sqrt 5} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 4 {\sqrt 5} \paren {\dfrac 1 {3 - \sqrt 5} } \times \paren {\dfrac {3 + \sqrt 5} {3 + \sqrt 5} } - \dfrac 4 {\sqrt 5} \paren {\dfrac 1 {3 + \sqrt 5} } \times \paren {\dfrac {3 - \sqrt 5} {3 - \sqrt 5} }\) | multiplying top and bottom by $\paren {3 + \sqrt 5}$ and by $\paren {3 - \sqrt 5}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 4 {\sqrt 5} \paren {\dfrac {3 + \sqrt 5} 4} - \dfrac 4 {\sqrt 5} \paren {\dfrac {3 - \sqrt 5} 4}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\dfrac {3 + \sqrt 5} {\sqrt 5} } - \paren {\dfrac {3 - \sqrt 5} {\sqrt 5} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2\) |
$\blacksquare$