与えられた数式の値を計算します。数式は $\frac{1}{2} \cdot 2^{n-1} \{2(2^{n-1} - 1) + (2^{n-1} - 1) \cdot 2\}$ です。代数学指数式の計算数式の展開2025/6/31. 問題の内容与えられた数式の値を計算します。数式は 12⋅2n−1{2(2n−1−1)+(2n−1−1)⋅2}\frac{1}{2} \cdot 2^{n-1} \{2(2^{n-1} - 1) + (2^{n-1} - 1) \cdot 2\}21⋅2n−1{2(2n−1−1)+(2n−1−1)⋅2} です。2. 解き方の手順まず、中括弧の中を整理します。2(2n−1−1)+(2n−1−1)⋅2=2⋅2n−1−2+2⋅2n−1−2=4⋅2n−1−4=4(2n−1−1)2(2^{n-1} - 1) + (2^{n-1} - 1) \cdot 2 = 2 \cdot 2^{n-1} - 2 + 2 \cdot 2^{n-1} - 2 = 4 \cdot 2^{n-1} - 4 = 4(2^{n-1} - 1)2(2n−1−1)+(2n−1−1)⋅2=2⋅2n−1−2+2⋅2n−1−2=4⋅2n−1−4=4(2n−1−1)次に、全体の式に代入します。12⋅2n−1{4(2n−1−1)}=12⋅2n−1⋅4(2n−1−1)=2⋅2n−1(2n−1−1)=2n(2n−1−1)=2n⋅2n−1−2n=22n−1−2n\frac{1}{2} \cdot 2^{n-1} \{4(2^{n-1} - 1)\} = \frac{1}{2} \cdot 2^{n-1} \cdot 4(2^{n-1} - 1) = 2 \cdot 2^{n-1} (2^{n-1} - 1) = 2^n (2^{n-1} - 1) = 2^n \cdot 2^{n-1} - 2^n = 2^{2n-1} - 2^n21⋅2n−1{4(2n−1−1)}=21⋅2n−1⋅4(2n−1−1)=2⋅2n−1(2n−1−1)=2n(2n−1−1)=2n⋅2n−1−2n=22n−1−2n3. 最終的な答え22n−1−2n2^{2n-1} - 2^n22n−1−2n