1. Use the root equation to get a recurrence

Since $\alpha, \beta$ are roots of $x^2-x-1=0,$ each root satisfies $r^2=r+1.$ Multiplying by $r^{n-2}$, we get for any integer $n\ge 2$: $r^n=r^{n-1}+r^{n-2}.$ So both $\alpha^n$ and $\beta^n$ satisfy the Fibonacci-type recurrence.

2. Apply this to $S_n$

Given $S_n=2023\alpha^n+2024\beta^n,$ we have \begin{align*} S_n &= 2023\alpha^n+2024\beta^n \ &= 2023(\alpha^{n-1}+\alpha^{n-2})+2024(\beta^{n-1}+\beta^{n-2}) \ &= (2023\alpha^{n-1}+2024\beta^{n-1})+(2023\alpha^{n-2}+2024\beta^{n-2}) \ &= S_{n-1}+S_{n-2}. \end{align*} Thus, $S_n=S_{n-1}+S_{n-2}.$

3. Substitute $n=12$

Then $S_{12}=S_{11}+S_{10}.$ This matches option B.

4. Check the other options

• A: $2S_{12}=S_{11}+S_{10}$

  But since $S_{12}=S_{11}+S_{10}$, this would mean $2S_{12}=S_{12},$ which is false in general.

• C: $S_{11}=S_{10}+S_{12}$

  Rearranging gives $S_{11}-S_{12}=S_{10}$, not consistent with the recurrence.

• D: $2S_{11}=S_{12}+S_{10}$

  Using $S_{12}=S_{11}+S_{10}$, $S_{12}+S_{10}=S_{11}+2S_{10},$ so this would require $S_{11}=2S_{10}$, not true in general.

Therefore, the correct option is B.