《2025年名校题库七年级数学上册》

1. (郫都区期末)已知对于任意正整数$n$,设$f(n)= \frac {1}{1+2+3+... +n}$,则$f(1)+f(2)+f(3)+... +f(100)$的值为____.

答案： $\frac{200}{101}$

2. (新都区期末)先观察下列各式,再完成下列各题.
$\frac {1}{1×2}= 1-\frac {1}{2};\frac {1}{2×3}= \frac {1}{2}-\frac {1}{3};\frac {1}{3×4}= \frac {1}{3}-\frac {1}{4};\frac {1}{4×5}= \frac {1}{4}-\frac {1}{5}.$
(1)①请仿照上面各式的结构写出:$\frac {1}{5×6}= $____;
②$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{n(n+1)}= $____.(其中$n$为整数,且满足$n≥1$)
(2)运用以上方法思考:求$\frac {1}{4}+\frac {1}{12}+\frac {1}{24}+\frac {1}{40}+\frac {1}{60}+\frac {1}{84}+\frac {1}{112}+\frac {1}{144}$的值.

答案：
(1)①$\frac{1}{5}-\frac{1}{6}$②$\frac{n}{n+1}$
(2)解:原式$=\frac{1}{2}× (\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots +\frac{1}{56}+\frac{1}{72})$$=\frac{1}{2}× (\frac{1}{1× 2}+\frac{1}{2× 3}+\frac{1}{3× 4}+\cdots +\frac{1}{7× 8}+\frac{1}{8× 9})$$=\frac{1}{2}× (1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{7}-\frac{1}{8}+$$\frac{1}{8}-\frac{1}{9})$$=\frac{1}{2}× (1-\frac{1}{9})=\frac{4}{9}.$

3. (石室联中)观察算式:
$\frac {1}{1×2}= 1-\frac {1}{2}= \frac {1}{2}$;
$\frac {1}{1×2}+\frac {1}{2×3}= 1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}= \frac {2}{3}$;
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}= 1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}= \frac {3}{4}$.
按规律填空:
(1)$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}= $____;
(2)$\frac {1}{1×4}+\frac {1}{4×7}+\frac {1}{7×10}+... +\frac {1}{2020×2023}= $____.

答案：
(1)$\frac{4}{5}$
(2)$\frac{674}{2023}$

4. (树德实验)
(1)计算:$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{99×100}$;
(2)计算:$\frac {1}{2×4}+\frac {1}{4×6}+\frac {1}{6×8}+... +\frac {1}{2020×2022}$;
(3)计算:$2019\frac {1}{10}+2020\frac {1}{40}+2021\frac {1}{88}+2022\frac {1}{154}+2023\frac {1}{238}$.

答案： 解:
(1)$\frac{1}{1× 2}+\frac{1}{2× 3}+\frac{1}{3× 4}+\cdots +\frac{1}{99× 100}=1-$$\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$$=\frac{99}{100}.$
(2)$\frac{1}{2× 4}+\frac{1}{4× 6}+\frac{1}{6× 8}+\cdots +\frac{1}{2020× 2022}=\frac{1}{2}×$$(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}× (\frac{1}{4}-\frac{1}{6})+\frac{1}{2}× (\frac{1}{6}-\frac{1}{8})+$$\cdots +\frac{1}{2}× (\frac{1}{2020}-\frac{1}{2022})=\frac{1}{2}× (\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-$$\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots +\frac{1}{2020}-\frac{1}{2022})=\frac{1}{2}× (\frac{1}{2}-$$\frac{1}{2022})=\frac{1}{2}× \frac{1010}{2022}=\frac{505}{2022}.$
(3)$2019\frac{1}{10}+2020\frac{1}{40}+2021\frac{1}{88}+2022\frac{1}{154}+$$2023\frac{1}{238}=2019+\frac{1}{10}+2020+\frac{1}{40}+2021+\frac{1}{88}+$$2022+\frac{1}{154}+2023+\frac{1}{238}=(2019+2020+2021+$$2022+2023)+(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238})=$$2021× 5+(\frac{1}{2× 5}+\frac{1}{5× 8}+\frac{1}{8× 11}+\frac{1}{11× 14}+$$\frac{1}{14× 17})=10105+\frac{1}{3}× (\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-$$\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17})=10105+\frac{1}{3}× (\frac{1}{2}-$$\frac{1}{17})=10105+\frac{1}{3}× \frac{15}{34}=10105+\frac{5}{34}=10105\frac{5}{34}.$

5. (郫都区期末)认真观察下面的序列等式的变化,寻找“将一项拆成两项”的规律:
$\frac {1}{1×2}= 1-\frac {1}{2},\frac {1}{2×3}= \frac {1}{2}-\frac {1}{3},\frac {1}{3×4}= \frac {1}{3}-\frac {1}{4},\frac {1}{4×5}= \frac {1}{4}-\frac {1}{5},\frac {1}{5×6}= \frac {1}{5}-\frac {1}{6},...$
用上面的思路,解答下列问题:
(1)写出上面序列等式的第$n$个等式;
(2)计算:$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+\frac {1}{7×9}+... +\frac {1}{2021×2023}$;
(3)计算:$\frac {1}{1+2}+\frac {1}{1+2+3}+\frac {1}{1+2+3+4}+\frac {1}{1+2+3+4+5}+... +\frac {1}{1+2+3+... +2023}$.

答案： 解:
(1)第n个等式为$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$
(2)$\frac{1}{1× 3}+\frac{1}{3× 5}+\frac{1}{5× 7}+\frac{1}{7× 9}+\cdots +\frac{1}{2021× 2023}$$=\frac{1}{2}× (1-\frac{1}{3})+\frac{1}{2}× (\frac{1}{3}-\frac{1}{5})+\frac{1}{2}× (\frac{1}{5}-$$\frac{1}{7})+\cdots +\frac{1}{2}× (\frac{1}{2021}-\frac{1}{2023})=\frac{1}{2}× (1-\frac{1}{3}+$$\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{1}{2021}-\frac{1}{2023})=\frac{1}{2}× (1-$$\frac{1}{2023})=\frac{1011}{2023}.$
(3)$\because 1+2+3+\cdots +n=\frac{n(n+1)}{2},$$\therefore \frac{1}{1+2+\cdots +n}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1}),$$\therefore \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+$$\frac{1}{1+2+3+4+5}+\cdots +\frac{1}{1+2+3+\cdots +2023}$$=2× (\frac{1}{2}-\frac{1}{3})+2× (\frac{1}{3}-\frac{1}{4})+2× (\frac{1}{4}-\frac{1}{5})$$+\cdots +2× (\frac{1}{2023}-\frac{1}{2024})$$=2× (\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots +\frac{1}{2023}-$$\frac{1}{2024})=2× (\frac{1}{2}-\frac{1}{2024})=\frac{1011}{1012}.$