《2025年暑假总动员宁夏人民教育出版社七升八数学苏科版》

17. 阅读材料：
求$1 + 2 + 2^{2} + 2^{3} + 2^{4} + … + 2^{2025}$的值.
解：设$S = 1 + 2 + 2^{2} + 2^{3} + 2^{4} + … + 2^{2024} + 2^{2025}$, ①
将等式两边同时乘2,得
$2S = 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + … + 2^{2025} + 2^{2026}$, ②
②-①得$2S - S = 2^{2026} - 1$,
即$S = 2^{2026} - 1$,
故$1 + 2 + 2^{2} + 2^{3} + 2^{4} + … + 2^{2025} = 2^{2026} - 1$.
请你仿照此法计算：
(1)$1 + 2 + 2^{2} + 2^{3} + 2^{4} + … + 2^{10}$；
(2)$1 + 3 + 3^{2} + 3^{3} + 3^{4} + … + 3^{n}$(其中$n$为正整数).

答案：
(1) 设 $ S = 1 + 2 + 2^{2} + 2^{3} + 2^{4} + \cdots + 2^{10} $，①
将等式两边同时乘 2，得 $ 2S = 2 + 2^{2} + 2^{3} + 2^{4} + \cdots + 2^{10} + 2^{11} $，②
② - ①得 $ 2S - S = 2^{11} - 1 $，即 $ S = 2^{11} - 1 $，
故 $ 1 + 2 + 2^{2} + 2^{3} + 2^{4} + \cdots + 2^{10} = 2^{11} - 1 $。
(2) 设 $ S = 1 + 3 + 3^{2} + 3^{3} + 3^{4} + \cdots + 3^{n} $，①
将等式两边同时乘 3，得 $ 3S = 3 + 3^{2} + 3^{3} + 3^{4} + \cdots + 3^{n} + 3^{n + 1} $，②
② - ①得 $ 3S - S = 3^{n + 1} - 1 $，即 $ 2S = 3^{n + 1} - 1 $，
解得 $ S = \frac{3^{n + 1} - 1}{2} $，
故 $ 1 + 3 + 3^{2} + 3^{3} + 3^{4} + \cdots + 3^{n} = \frac{3^{n + 1} - 1}{2} $。

18. 观察下列等式：
$\frac{1}{1×2}= 1 - \frac{1}{2}$,$\frac{1}{2×3}= \frac{1}{2} - \frac{1}{3}$,$\frac{1}{3×4}= \frac{1}{3} - \frac{1}{4}$,$\frac{1}{4×5}= \frac{1}{4} - \frac{1}{5}$,….
(1)猜想并写出：$\frac{1}{n(n + 1)}=$$\frac{1}{n} - \frac{1}{n + 1}$；
(2)用你猜想的规律,计算：$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + … + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)}$.
$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \cdots + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)}$
$= \frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + \frac{1}{4×5} + \cdots + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \cdots + (\frac{1}{n - 1} - \frac{1}{n}) + (\frac{1}{n} - \frac{1}{n + 1})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots + \frac{1}{n - 1} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}$
$= 1 - \frac{1}{n + 1}$
$= \frac{n}{n + 1}$

答案：
(1) $\frac{1}{n} - \frac{1}{n + 1}$
(2) $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \cdots + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)}$
$= \frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + \frac{1}{4×5} + \cdots + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \cdots + (\frac{1}{n - 1} - \frac{1}{n}) + (\frac{1}{n} - \frac{1}{n + 1})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots + \frac{1}{n - 1} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}$
$= 1 - \frac{1}{n + 1}$
$= \frac{n}{n + 1}$