《2025年通成学典课时作业本七年级数学上册苏科版苏州专版》

4. 计算：
(1)$-1-2-3-4-... -199-200$； (2)$1+5+5^{2}+5^{3}+... +5^{217}+5^{218}$。

答案：
(1) 设 $S = -1 - 2 - 3 - 4 - \cdots - 199 - 200$ ①，则 $S = -200 - 199 - 198 - 197 - \cdots - 2 - 1$ ②. 由 ① + ②，得 $2S = -201 × 200$，即 $2S = -40200$，所以 $S = -20100$，即 $-1 - 2 - 3 - 4 - \cdots - 199 - 200 = -20100$
(2) 令 $S = 1 + 5 + 5^2 + 5^3 + \cdots + 5^{217} + 5^{218}$ ①，则 $5S = 5 + 5^2 + 5^3 + 5^4 + \cdots + 5^{218} + 5^{219}$ ②. 由 ② - ①，得 $5S - S = 5^{219} - 1$，所以 $S = \frac{5^{219} - 1}{4}$，即 $1 + 5 + 5^2 + 5^3 + \cdots + 5^{217} + 5^{218} = \frac{5^{219} - 1}{4}$

5. 计算：$1+\frac {1}{2}+\frac {1}{2^{2}}+\frac {1}{2^{3}}+... +\frac {1}{2^{2026}}$。

答案： 设 $S = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{2026}}$ ①，则 $\frac{1}{2}S = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \cdots + \frac{1}{2^{2027}}$ ②. 由 ① - ②，得 $\frac{1}{2}S = 1 - \frac{1}{2^{2027}}$，所以 $S = 2 - \frac{1}{2^{2026}}$，即 $1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{2026}} = 2 - \frac{1}{2^{2026}}$

6. 阅读材料：
在计算$\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+... +\frac {1}{420}$时,直接计算很繁琐,我们可以采用“裂项——消项”法简化运算。
$\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+... +\frac {1}{420}= \frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+... +\frac {1}{20×21}= (1-\frac {1}{2})+(\frac {1}{2}-\frac {1}{3})+(\frac {1}{3}-\frac {1}{4})+(\frac {1}{4}-\frac {1}{5})+... +(\frac {1}{20}-\frac {1}{21})= 1-\frac {1}{21}= \frac {20}{21}$。
方法应用：
试用“裂项——消项”法解下面各题：
(1)$\frac {1}{3×7}+\frac {1}{7×11}+\frac {1}{11×15}+... +\frac {1}{55×59}$； (2)$-\frac {1}{3}-\frac {1}{15}-\frac {1}{35}-\frac {1}{63}-\frac {1}{99}-\frac {1}{143}$。

答案：
(1) 原式 $= \frac{1}{4} × (\frac{1}{3} - \frac{1}{7}) + \frac{1}{4} × (\frac{1}{7} - \frac{1}{11}) + \frac{1}{4} × (\frac{1}{11} - \frac{1}{15}) + \cdots + \frac{1}{4} × (\frac{1}{55} - \frac{1}{59}) = \frac{1}{4} × (\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \frac{1}{11} - \frac{1}{15} + \cdots + \frac{1}{55} - \frac{1}{59}) = \frac{1}{4} × (\frac{1}{3} - \frac{1}{59}) = \frac{14}{177}$
(2) 原式 $= -\frac{1}{1 × 3} - \frac{1}{3 × 5} - \frac{1}{5 × 7} - \frac{1}{7 × 9} - \frac{1}{9 × 11} - \frac{1}{11 × 13} = -\frac{1}{2} × (1 - \frac{1}{3}) - \frac{1}{2} × (\frac{1}{3} - \frac{1}{5}) - \frac{1}{2} × (\frac{1}{5} - \frac{1}{7}) - \frac{1}{2} × (\frac{1}{7} - \frac{1}{9}) - \frac{1}{2} × (\frac{1}{9} - \frac{1}{11}) - \frac{1}{2} × (\frac{1}{11} - \frac{1}{13}) = -\frac{1}{2} × (1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{11} + \frac{1}{11} - \frac{1}{13}) = -\frac{1}{2} × (1 - \frac{1}{13}) = -\frac{6}{13}$