Animated Solution for Mathematics - Quadratic Equations: If $\alpha ,\beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023{\alpha }^n+2024{\beta }^n$, then

Welcome, future engineers. Today, we are not just solving a problem; we are uncovering a hidden rhythm in mathematics. When you look at the equation $x^2-x-1=0$, what do you see?

Perhaps you see a simple quadratic. But I want you to see a gateway. This specific equation is the foundation of the Golden Ratio, a number that appears in the spirals of galaxies and the petals of flowers.

Our task is to analyze the sequence $S_n=2023{\alpha }^n+2024{\beta }^n$. At first glance, this looks like a calculation nightmare. Let us find the path of least resistance.

Let us visualize the function $f(x)=x^2-x-1$. This is a parabola opening upwards, and the roots $\alpha$ and $\beta$ are the points where this parabola kisses the $x$-axis.

By definition, if $\alpha$ is a root, it must satisfy the equation. Therefore:

This is our golden key. Rearranging this, we get ${\alpha }^2=\alpha +1$. This simple identity is the secret to avoiding all the messy arithmetic.

It tells us that any power of $\alpha$ can be reduced to a lower power. If we have ${\alpha }^3$, we can write it as $\alpha ({\alpha }^2)=\alpha (\alpha +1)={\alpha }^2+\alpha$. We can keep reducing until we have only linear terms.

The same logic applies perfectly to $\beta$. We have ${\beta }^2=\beta +1$. Keep these two relations etched in your mind.

Now, look at our sequence $S_n=2023{\alpha }^n+2024{\beta }^n$. We want to relate $S_{12}$, $S_{11}$, and $S_{10}$. Let us test the hypothesis that the sequence follows a recurrence relation by calculating the sum of the two previous terms:

Now, watch the grouping. We group the $\alpha$ terms and the $\beta$ terms:

This is where the magic happens. Factor out the lowest power, which is ${\alpha }^{n-2}$ and ${\beta }^{n-2}$:

Remember our golden key? $\alpha +1={\alpha }^2$ and $\beta +1={\beta }^2$. Let us substitute these back into our expression:

Using the laws of exponents, ${\alpha }^{n-2}\cdot {\alpha }^2={\alpha }^n$. The expression becomes:

And look at that! We have arrived exactly back at $S_n$. We have proven that $S_n=S_{n-1}+S_{n-2}$.

This is a linear recurrence relation. It is robust, elegant, and completely independent of the specific values of the roots. For $n=12$, the relation becomes $S_{12}=S_{11}+S_{10}$.

You see, mathematics is not about brute force; it is about finding the underlying structure. When you stop fighting the numbers and start listening to the patterns, the problems solve themselves.