Finding a Fibonacci recurrence for linear combinations of roots

Key concept: This problem uses a key fact about quadratic roots: any sequence built from their powers satisfies the same recurrence as the roots.

Step by step

Key idea

Let the roots of $x^2-x-1=0$ be $\alpha$ and $\beta$. Since each root satisfies

$r^2=r+1,$

any power sequence $r^n$ obeys the Fibonacci recurrence

$r^n=r^{n-1}+r^{n-2}\quad (n\ge 2).$

So the sequence

$S_n=2025\alpha^n-2026\beta^n$

also satisfies

$S_n=S_{n-1}+S_{n-2}.$

That means the target quantity can be rewritten as

$S_{10}+S_{11}=S_{12}.$

Find a cleaner closed form

Instead of computing many terms, use the special identities for the roots. From $x^2-x-1=0$, we have

$\alpha+\beta=1,\qquad \alpha\beta=-1.$

A very efficient way is to express $S_n$ in terms of the standard sequences generated by $\alpha^n$ and $\beta^n$. Let

$T_n=\alpha^n-\beta^n.$

Then $T_n$ also satisfies the same recurrence. We can write

$S_n=2025(\alpha^n-\beta^n)-\beta^n = 2025T_n-\beta^n,$

but this is not yet the cleanest route.

A better observation is that the recurrence determines $S_{12}$ from two initial values. Compute

$S_0=2025-2026=-1,$

and

$S_1=2025\alpha-2026\beta=2025(\alpha-\beta)-\beta.$

Since

$\alpha-\beta=\sqrt{5},\qquad \beta=\frac{1-\sqrt5}{2},$

we get

$S_1=2025\sqrt5-\frac{1-\sqrt5}{2}=\frac{4051\sqrt5-1}{2}.$

Now use the recurrence repeatedly. The sequence follows the Fibonacci pattern:

$S_n=F_{n-1}S_1+F_{n-2}S_0,$

for $n\ge 2$, where $F_0=0,F_1=1$. Thus

$S_{12}=F_{11}S_1+F_{10}S_0=89S_1+55(-1).$

Substitute $S_1$:

=\frac{360539\sqrt5-89}{2}-55 =\frac{360539\sqrt5-199}{2}.$$ Therefore, $$\boxed{S_{10}+S_{11}=\frac{360539\sqrt5-199}{2}}.$$ ## Common recurrence shortcut Because $S_n=S_{n-1}+S_{n-2}$, you do not need to build $S_2,S_3,\dots$ one by one. Once you know $S_0$ and $S_1$, every later term is a Fibonacci-weighted combination of those two starting values. That is the fastest method for this type of problem. ### Pitfall alert A common mistake is to compute $S_1$ and $S_2$ in a messy way and then continue the recurrence by hand. That works, but it is easy to lose track of signs and coefficients. Another frequent error is forgetting that $S_{10}+S_{11}=S_{12}$ only because the sequence satisfies the Fibonacci recurrence; this identity does not hold for arbitrary sequences. Also, when using $\alpha-\beta=\sqrt5$, make sure the root ordering is consistent with $\alpha>\beta$, otherwise the sign flips and the final answer changes. ### Try different conditions If the problem instead asked for $S_{10}$ alone, the same method would still work: use $S_n=F_{n-1}S_1+F_{n-2}S_0$ and plug in $n=10$. If the recurrence were changed to $x^2-ax-b=0$, then the same idea would produce $S_n=aS_{n-1}+bS_{n-2}$, but the Fibonacci coefficients would be replaced by the corresponding characteristic-sequence coefficients. For example, with $S_n=2025\alpha^n-2026\beta^n$ and roots of $x^2-3x+1=0$, the same structural method applies, but the closed form and recurrence constants would differ. ### Further reading characteristic equation, Fibonacci recurrence, Vieta's formulas