已知数列 {an}\left\{ {a}_{n}\right\}{an} 满足 a1=1{a}_{1} = 1a1=1 ,前 nnn 项和为 Sn{S}_{n}Sn , an+1⋅an=2n(n∈N∗){a}_{n + 1} \cdot {a}_{n} = {2}^{n}\left( {n \in {\mathbf{N}}^{ * }}\right)an+1⋅an=2n(n∈N∗) ,则 S2024{S}_{2024}S2024 等于（ ）A. 22024−1{2}^{2024} - 122024−1B. 3×21012−13 \times {2}^{1012} - 13×21012−1C. 3×21012−23 \times {2}^{1012} - 23×21012−2D. 3×21012−33 \times {2}^{1012} - 33×21012−3