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第33页

信息发布者：

$a^{2}-b^{2}=(a + b)(a - b)$

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精灵详解）

解：$(3)100^2\pi -99^2\pi +… + 4^2\pi -3^2\pi +2^2\pi -1^2\pi$

$=\pi (100^2-99^2+… + 4^2-3^2+2^2-1^2)$

$=\pi (100 + 99+… + 4 + 3 + 2 + 1)$

$=\pi ×\frac {100×(100 + 1)}{2}$

$=5050\pi (\mathrm {cm}^2)$

故所有阴影的面积和为$5050\pi\mathrm {cm}^2。$

解：原式$=(3 - 1)×(3 + 1)×(3^2+1)×(3^4+1)×$

$(3^8+1)+1$

$ =(3^2-1)×(3^2+1)×(3^4+1)×(3^8+1)+1$

$ =(3^4-1)×(3^4+1)×(3^8+1)+1$

$ =(3^8-1)×(3^8+1)+1$

$ =3^{16}$

解：

$\begin {aligned}&(1-\frac {1}{2^2})(1-\frac {1}{3^2})(1-\frac {1}{4^2})(1-\frac {1}{5^2})…(1-\frac {1}{2023^2})(1-\frac {1}{2024^2})\\=&(1+\frac {1}{2})(1-\frac {1}{2})(1+\frac {1}{3})(1-\frac {1}{3})(1+\frac {1}{4})(1-\frac {1}{4})…(1+\frac {1}{2023})(1-\frac {1}{2023})(1+\frac {1}{2024})(1-\frac {1}{2024})\\=&(1+\frac {1}{2})(1+\frac {1}{3})(1+\frac {1}{4})…(1+\frac {1}{2023})(1+\frac {1}{2024})(1-\frac {1}{2})(1-\frac {1}{3})(1-\frac {1}{4})…(1-\frac {1}{2023})(1-\frac {1}{2024})\\=&\frac {3}{2}×\frac {4}{3}×\frac {5}{4}×…×\frac {2024}{2023}×\frac {2025}{2024}×\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×…×\frac {2022}{2023}×\frac {2023}{2024}\\=&\frac {2025}{2}×\frac {1}{2024}\\=&\frac {2025}{4048}\end {aligned}$