Proof $\displaystyle \sum_{k=0}^n \lfloor \varphi^k\rfloor=\left\lfloor\dfrac{\varphi^{n+1}-1}{\varphi-1}-\dfrac{n-1+(-1)^n}{2}\right\rfloor$

Question

When I was trying approximate this sum $\displaystyle A_n=\sum_{k=0}^n\left\lfloor\varphi^k\right\rfloor$, where $\varphi=\dfrac{1+\sqrt 5}{2}$ is the golden-ratio, I found that $B_n=\left\lfloor\dfrac{\varphi^{n+1}-1}{\varphi-1}-\dfrac{n-1+(-1)^n}{2}\right\rfloor$ is exactly $A_n$ for all $n\ge 2$. Can you help me proving that $A_n= B_n$?