Partial Sums of Power Series with Fibonacci Coefficients

Theorem

$\ds \sum_{k \mathop = 0}^n F_k x^k = \begin {cases} \dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\ \dfrac {\paren {n + 1} x^n F_{n + 1} + \paren {n + 2} x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end {cases}$

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

Proof

Multiplying the summation by $x^2 + x - 1$:

$\ds $

$\ds \sum_{k \mathop = 0}^n F_k x^k \paren {x^2 + x - 1}$

$\ds $

$=$

$\ds F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots + F_{n - 2} x^n + F_{n - 1} x^{n + 1} + F_n x^{n + 2}$

$\ds $

$\, \ds + \, $

$\ds F_0 x + F_1 x^2 + F_2 x^3 + \cdots + F_{n - 2} x^{n - 1} + F_{n - 1} x^n + F_n x^{n + 1}$

$\ds $

$\, \ds - \, $

$\ds F_0 - F_1 x - F_2 x^2 - \cdots - F_{n - 2} x^{n - 2} - F_{n - 1} x^{n - 1} - F_n x^n$

$\ds $

$=$

$\ds F_0 + \paren {F_0 - F_1} x + \paren {F_{n - 1} + F_n} x^{n + 1} + F_n x^{n + 2}$

as, in general, $\paren {F_{j - 2} + F_{j - 1} - F_j} x^j = 0$ by Definition of Fibonacci Number

$\ds $

$=$

$\ds 0 + \paren {-1} x + F_{n + 1} x^{n + 1} + F_n x^{n + 2}$

Definition of Fibonacci Number: $F_0 = 0, F_1 = 1$

$\ds \leadsto \ \ $

$\ds \sum_{k \mathop = 0}^n F_k x^k$

$=$

$\ds \dfrac {F_{n + 1} x^{n + 1} + F_n x^{n + 2} - x} {x^2 + x - 1}$

If the denominator is $0$, then $x = \dfrac 1 \phi$ or $x = \dfrac 1 {\hat \phi}$ and the numerator is $0$ also.

Thus we can differentiate the numerator and denominator with respect to $x$ and use L'Hôpital's Rule:

$\ds $

$\ds \dfrac {\map {\dfrac \d {\d x} } {F_{n + 1} x^{n + 1} + F_n x^{n + 2} - x} } {\map {\dfrac \d {\d x} } {x^2 + x - 1} }$

$\ds $

$=$

$\ds \dfrac {\paren {n + 1} F_{n + 1} x^n + \paren {n + 2} F_n x^{n + 1} - 1} {2 x + 1}$

Power Rule for Derivatives

Hence the result.

$\blacksquare$

Sources

1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $21$