Partial Sums of Power Series with Fibonacci Coefficients

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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 $x^2 + x - 1$:

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


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:

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


Hence the result.

$\blacksquare$


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