Bulgarian Winter Tournament

Since $$2A_n=1\cdot 4+3\cdot 8+\cdots +(2n-3)\cdot 2^n+(2n-1)\cdot 2^{n+1},$$ it follows $$\begin{array}{rl}{A}_n=2A_n-A_n& =(2n-1)\cdot 2^{n+1}-(1\cdot 2+2\cdot 4+2\cdot 8+\cdots +2\cdot 2^n)\\ & =(2n-1)\cdot 2^{n+1}-2\cdot (2+4+8+\cdots +2^n)+1\cdot 2\\ & =(2n-1)\cdot 2^{n+1}-2\cdot 2\cdot \frac{2^n-1}{2-1}+2=(2n-3)\cdot 2^{n+1}+6.\end{array}$$ Therefore $A_{2024}=4045\cdot 2^{2025}+6$.