《2025年学霸题中题七年级数学上册苏科版》

8. 计算:
(1)$\frac{1}{2024} + \frac{2}{2024} + \frac{3}{2024} + … + \frac{4047}{2024}$;
(2)$\frac{1}{2} - (\frac{1}{3} + \frac{2}{3}) + (\frac{1}{4} + \frac{2}{4} + \frac{3}{4}) - … + (\frac{1}{2024} + \frac{2}{2024} + … + \frac{2023}{2024})$.

答案：
(1) 设 $ A = \frac{1}{2024} + \frac{2}{2024} + \frac{3}{2024} + \cdots + \frac{4047}{2024} $，$ B = \frac{4047}{2024} + \frac{4046}{2024} + \cdots + \frac{1}{2024} $，则 $ A = B $，$ A + B = 2A = \left( \frac{1}{2024} + \frac{4047}{2024} \right) + \left( \frac{2}{2024} + \frac{4046}{2024} \right) + \cdots + \left( \frac{4046}{2024} + \frac{2}{2024} \right) + \left( \frac{4047}{2024} + \frac{1}{2024} \right) = \frac{4048}{2024} + \frac{4048}{2024} + \cdots + \frac{4048}{2024} = 2 × 4047 = 8094 $，所以 $ A = 4047 $，即原式 $ = 4047 $。
(2) 设 $ B = \frac{1}{2} - \left( \frac{1}{3} + \frac{2}{3} \right) + \left( \frac{1}{4} + \frac{2}{4} + \frac{3}{4} \right) - \cdots + \left( \frac{1}{2024} + \frac{2}{2024} + \cdots + \frac{2023}{2024} \right) $ ①，所以 $ B = \frac{1}{2} - \left( \frac{2}{3} + \frac{1}{3} \right) + \left( \frac{3}{4} + \frac{2}{4} + \frac{1}{4} \right) - \cdots + \left( \frac{2023}{2024} + \frac{2022}{2024} + \cdots + \frac{1}{2024} \right) $ ②，① + ② 得 $ 2B = 1 - (1 + 1) + (1 + 1 + 1) - \cdots + (1 + 1 + \cdots + 1) = 1 - 2 + 3 - \cdots - 2022 + 2023 = -1011 + 2023 = 1012 $，所以 $ B = 506 $，即原式 $ = 506 $。

9. 计算$333×999 - 1002×332= $3.

答案： 3 解析: $ 333×999 = 333×(1000 - 1) = 333×1000 - 333 $，$ 1002×332 = (1000 + 2)×332 = 332×1000 + 664 $，所以 $ 333×999 - 1002×332 = 333×1000 - 332×1000 - 333 - 664 = 1000 - 333 - 664 = 3 $。

10. 计算:$1 + 3\frac{1}{6} + 5\frac{1}{12} + 7\frac{1}{20} + 9\frac{1}{30} + 11\frac{1}{42} + 13\frac{1}{56} + 15\frac{1}{72} + 17\frac{1}{90}$.

答案： 原式 $ = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) + \left( \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} \right) = 81 + \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{10} \right) = 81 + \left( \frac{1}{2} - \frac{1}{10} \right) = 81 + \frac{2}{5} = 81\frac{2}{5} $。

11. 计算:
(1)$\frac{1}{2} + (\frac{1}{2})^{2} + (\frac{1}{2})^{3} + (\frac{1}{2})^{4} + … + (\frac{1}{2})^{8}$;
(2)$5 + 2×5^{2} + 3×5^{3} + 4×5^{4} + … + 8×5^{8}$.

答案：
(1) 设 $ S = \frac{1}{2} + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^4 + \cdots + \left( \frac{1}{2} \right)^8 $ ①，则 $ 2S = 1 + \frac{1}{2} + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^3 + \cdots + \left( \frac{1}{2} \right)^7 $ ②，则 ② - ①，得 $ 2S - S = S = 1 - \frac{1}{2^8} = \frac{255}{256} $。
(2) 设 $ M = 5 + 2×5^2 + 3×5^3 + 4×5^4 + \cdots + 8×5^8 $ ①，所以 $ 5M = 1×5^2 + 2×5^3 + 3×5^4 + \cdots + 8×5^9 $ ②，所以 ② - ① 得 $ 5M - M = (1×5^2 + 2×5^3 + 3×5^4 + \cdots + 8×5^9) - (5 + 2×5^2 + 3×5^3 + 4×5^4 + \cdots + 8×5^8) = 8×5^9 - (5 + 5^2 + 5^3 + \cdots + 5^8) $，设 $ T = 5 + 5^2 + 5^3 + \cdots + 5^8 $，所以 $ 5T = 5^2 + 5^3 + \cdots + 5^9 $，所以 $ 5T - T = 5^9 - 5 $，所以 $ 4T = 5^9 - 5 $，所以 $ T = \frac{5^9 - 5}{4} $，所以 $ 5M - M = 8×5^9 - \frac{5^9 - 5}{4} $，所以 $ 4M = 8×5^9 - \frac{5^9 - 5}{4} $，所以 $ M = \frac{8×5^9}{4} - \frac{5^9 - 5}{16} = \frac{31×5^9 + 5}{16} $。

12. 计算:
(1)$(\frac{1}{6} + \frac{1}{7} + \frac{1}{8}) - 4×(\frac{1}{2} - \frac{1}{6} - \frac{1}{7} - \frac{1}{8}) - 5×(\frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9})= $$-1\frac{4}{9}$;
(2)$(\frac{1}{2} + \frac{1}{3} + … + \frac{1}{2025})×(1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{2024}) - (1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{2025})×(\frac{1}{2} + \frac{1}{3} + … + \frac{1}{2024})= $$\frac{1}{2025}$.

答案：
(1)$-1\frac{4}{9}$ 解析:把 $ \frac{1}{6} + \frac{1}{7} + \frac{1}{8} $ 当成一个整体，则原式 $ = \left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) - 4×\left[ \frac{1}{2} - \left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \right] - 5×\left[ \left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) - \frac{1}{9} \right] = \left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) - 2 + 4×\left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) - 5×\left( \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \frac{5}{9} = -2 + \frac{5}{9} = -1\frac{4}{9} $。
(2)$\frac{1}{2025}$ 解析:设 $ a = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2025} $，$ b = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2024} $，则原式 $ = a(1 + b) - (1 + a)b = a + ab - b - ab = a - b = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2025} - \left( \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2024} \right) = \frac{1}{2025} $。