Sum of nth Fibonacci Number over nth Power of 2/Proof 3

Theorem

$\ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n} = 2$

where $F_n$ is the $n$th Fibonacci number.

Proof

\(\ds \sum_{k \mathop = 0}^{\infty} F_k z^k\)

\(=\)

\(\ds \dfrac z {1 - z - z^2}\)

Generating Function for Fibonacci Numbers

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^{\infty} \frac {F_k} {2^k}\)

\(=\)

\(\ds \dfrac {\dfrac 1 2} {1 - \dfrac 1 2 - \paren {\dfrac 1 2}^2}\)

substituting $z = \dfrac 1 2$

\(\ds \)

\(=\)

\(\ds \dfrac {1 / 2} {1 / 4}\)

\(\ds \)

\(=\)

\(\ds 2\)

$\blacksquare$