I am a student curious about recurrence relations who has just bumped into the Intermediate 1st year (grade 11).
I derived the closed form expressions for the $n^{th}$ term and the sum of $n$ terms of a Fibonacci-like sequence of the form $(a,b,a+b,a+2b,2a+3b,...)$ by substituting $a_n=x^n$ into the characteristic recurrence relation: $a_{n+2}=a_{n+1}+a_n$.
I want to know what this method ($a_n=x^n$) is called, and if the below mentioned expressions are already documented somewhere. I would also like to know if there are simpler versions of these expressions.
$$T_n=\sum_{r \in \{{\phi,\psi\}}}{r^n \left(\frac{T_1}{3r+1}+\frac{T_2}{r+2}\right)}$$
$$S_n=\left(\sum_{r\in\{\phi,\psi\}}{r^n\left(\frac{T_1+T_2}{3r+1}+\frac{T_1+2T_2}{r+2}\right)}\right)-T_2$$
where,
$T_n = n^{th}$ term of the Fibonacci-like sequence.
$S_n =$ sum of $n$ terms of the Fibonacci-like sequence.
$\phi,\psi$ are the roots of the equation: $x^2-x-1=0$
Also, I saw other answers mentioning the closed form expressions for $T_n$ using Binet's formula and other methods, but I could not find a closed form expression for $S_n$ in any of those.
These work over every Fibonacci-like sequence. I would love suggestions on the improvement of these expressions.
Thank You.
EDIT: There seems to be some confusion around the summation in the comments. Here is an example that completely explains it.
Let's consider the Fibonacci sequence:$$1,1,2,3,5,8,...$$
Let's verify the formulas for the 6th term and the sum of the first 6 terms of the Fibonacci sequence using the expressions provided above.
First, let's calculate the 6th term ($T_6$) of the Fibonacci sequence:
$$T_6 = \sum_{r \in \{\phi,\psi\}}{r^6\left(\frac{T_1}{3r+1}+\frac{T_2}{r+2}\right)}$$
We know the first two terms of the Fibonacci sequence are $T_1 = 1$ and $T_2 = 1$, and we'll use the roots of the characteristic equation, $\phi$ and $\psi$, which are approximately $\phi \approx 1.61803$ and $\psi \approx -0.61803$.
Substituting these values into the formula:
$$T_6 = \left(\phi^6\left(\frac{1}{3\phi+1}+\frac{1}{\phi+2}\right)\right) + \left(\psi^6\left(\frac{1}{3\psi+1}+\frac{1}{\psi+2}\right)\right)$$
After evaluating this expression, we get $T_6 \approx 8$.
Next, let's calculate the sum of the first 6 terms of the Fibonacci sequence ($S_6$):
$$S_6 = \left(\sum_{r \in \{\phi,\psi\}}{r^6\left(\frac{T_1+T_2}{3r+1}+\frac{T_1+2T_2}{r+2}\right)}\right)-T_2$$
Substituting the known values:
$$S_6 = \left(\phi^6\left(\frac{1+1}{3\phi+1}+\frac{1+2}{\phi+2}\right)\right) + \left(\psi^6\left(\frac{1+1}{3\psi+1}+\frac{1+2}{\psi+2}\right)\right) - 1$$
After evaluating this expression, we get $S_6 \approx 20$.
The values obtained from the formulas match the actual values of the 6th term and the sum of the first 6 terms of the Fibonacci sequence, confirming the correctness of the expressions provided above.