1. 计算 $3^{2024}-5×3^{2023}+6×3^{2022}$ 的结果是 (
A
)
A.0
B.$3^{1008}$
C.$3^{2010}$
D.$3^{2011}$
答案：1.A 点拨:原式$=3^{2022+2}-5×3^{2022+1}+6×3^{2022}=3^{2022}×3^{2}-5×3^{2022}×3+6×3^{2022}=3^{2022}×(3^{2}-5×3+6)=3^{2022}×(9-15+6)=0.$
2. 计算 $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-2×(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5})-3×(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6})$ 的结果为
$-\frac{1}{2}$
.
答案：2.$-\frac {1}{2}$ 点拨:设$a=\frac {1}{3}+\frac {1}{4}+\frac {1}{5}$,则原式$=a-2(\frac {1}{2}-a)-3(a-\frac {1}{6})=a-1+2a-3a+\frac {1}{2}=-\frac {1}{2}.$
解析：
解：设$a = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$，则原式可化为：
$\begin{aligned}&a - 2\left(\frac{1}{2} - a\right) - 3\left(a - \frac{1}{6}\right)\\=&a - 2×\frac{1}{2} + 2a - 3a + 3×\frac{1}{6}\\=&a - 1 + 2a - 3a + \frac{1}{2}\\=&(a + 2a - 3a) + \left(-1 + \frac{1}{2}\right)\\=&0 - \frac{1}{2}\\=&-\frac{1}{2}\end{aligned}$
$-\frac{1}{2}$
3. 计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2024×2025}= \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2024}-\frac{1}{2025}= 1-\frac{1}{2025}= \frac{2024}{2025}$. 观察上述解题方法,思考其中的规律,并解答下列问题：
(1)计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{n(n+1)}= $
$\frac{n}{n+1}$
；(直接写结果)
(2)计算：$\frac{1}{51×52}+\frac{1}{52×53}+\frac{1}{53×54}+…+\frac{1}{2024×2025}$；
解:原式$=\frac {1}{1×2}+\frac {1}{2×3}+... +\frac {1}{50×51}+\frac {1}{51×52}+\frac {1}{52×53}+\frac {1}{53×54}+... +\frac {1}{2024×2025}-(\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{50×51})=\frac {2024}{2025}-\frac {50}{51}=\frac {658}{34425}.$
(3)计算：$\frac{1}{1×3}+\frac{1}{3×5}+\frac{1}{5×7}+…+\frac{1}{2023×2025}$；
解:原式$=\frac {1}{2}×(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+... +\frac {1}{2023}-\frac {1}{2025})=\frac {1}{2}×(1-\frac {1}{2025})=\frac {1}{2}×\frac {2024}{2025}=\frac {1012}{2025}.$
(4)计算：$1-\frac{1}{1×4}-\frac{1}{4×7}-\frac{1}{7×10}-…-\frac{1}{31×34}$.
解:原式$=1-\frac {1}{3}×(1-\frac {1}{4}+\frac {1}{4}-\frac {1}{7}+\frac {1}{7}-\frac {1}{10}+... +\frac {1}{31}-\frac {1}{34})=1-\frac {1}{3}×(1-\frac {1}{34})=1-\frac {1}{3}×\frac {33}{34}=1-\frac {11}{34}=\frac {23}{34}.$
答案：3.(1)$\frac {n}{n+1}$
(2)解:原式$=\frac {1}{1×2}+\frac {1}{2×3}+... +\frac {1}{50×51}+\frac {1}{51×52}+\frac {1}{52×53}+\frac {1}{53×54}+... +\frac {1}{2024×2025}-(\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{50×51})=\frac {2024}{2025}-\frac {50}{51}=\frac {658}{34425}.$
(3)解:原式$=\frac {1}{2}×(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+... +\frac {1}{2023}-\frac {1}{2025})=\frac {1}{2}×(1-\frac {1}{2025})=\frac {1}{2}×\frac {2024}{2025}=\frac {1012}{2025}.$
(4)解:原式$=1-\frac {1}{3}×(1-\frac {1}{4}+\frac {1}{4}-\frac {1}{7}+\frac {1}{7}-\frac {1}{10}+... +\frac {1}{31}-\frac {1}{34})=1-\frac {1}{3}×(1-\frac {1}{34})=1-\frac {1}{3}×\frac {33}{34}=1-\frac {11}{34}=\frac {23}{34}.$
