Partial Sums of Power Series with Fibonacci Coefficients

Theorem

$\displaystyle \sum_{k \mathop = 0}^n F_k x^k = \begin{cases} \dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\ \dfrac {\left({n + 1}\right) x^n F_{n + 1} + \left({n + 2}\right) x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end{cases}$

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

Proof

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

 $\ds$  $\ds \sum_{k \mathop = 0}^n F_k x^k \left({x^2 + x - 1}\right)$ $\ds$ $=$ $\ds F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots + F_{n - 2} x^n + F_{n - 1} x^{n + 1} + F_n x^{n + 2}$ $\ds$  $\, \ds + \,$ $\ds F_0 x + F_1 x^2 + F_2 x^3 + \cdots + F_{n - 2} x^{n - 1} + F_{n - 1} x^n + F_n x^{n + 1}$ $\ds$  $\, \ds - \,$ $\ds F_0 - F_1 x - F_2 x^2 - \cdots - F_{n - 2} x^{n - 2} - F_{n - 1} x^{n - 1} - F_n x^n$ $\ds$ $=$ $\ds F_0 + \left({F_0 - F_1}\right) x + \left({F_{n - 1} + F_n}\right) x^{n + 1} + F_n x^{n + 2}$ as, in general, $\left({F_{j - 2} + F_{j - 1} - F_j}\right) x^j = 0$ by Definition of Fibonacci Number $\ds$ $=$ $\ds 0 + \left({-1}\right) x + F_{n + 1} x^{n + 1} + F_n x^{n + 2}$ 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:

 $\ds$  $\ds \dfrac {\dfrac \d {\d x} \left({F_{n + 1} x^{n + 1} + F_n x^{n + 2} - x}\right)} {\dfrac \d {\d x} \left({x^2 + x - 1}\right)}$ $\ds$ $=$ $\ds \dfrac {\left({n + 1}\right) F_{n + 1} x^n + \left({n + 2}\right) F_n x^{n + 1} - 1} {2 x + 1}$ Power Rule for Derivatives

Hence the result.

$\blacksquare$