《2025年名校题库七年级数学上册》

13. (七中育才)计算:$\frac {1}{2}+(\frac {1}{3}+\frac {2}{3})+(\frac {1}{4}+\frac {2}{4}+\frac {3}{4})+... +(\frac {1}{60}+\frac {2}{60}+... +\frac {59}{60})$.

答案： 885

14. (成外)观察下列各式的计算结果:
$1-\frac {1}{2^{2}}= 1-\frac {1}{4}= \frac {3}{4}= \frac {1}{2}×\frac {3}{2}$;
$1-\frac {1}{3^{2}}= 1-\frac {1}{9}= \frac {8}{9}= \frac {2}{3}×\frac {4}{3}$;
$1-\frac {1}{4^{2}}= 1-\frac {1}{16}= \frac {15}{16}= \frac {3}{4}×\frac {5}{4}$;
$1-\frac {1}{5^{2}}= 1-\frac {1}{25}= \frac {24}{25}= \frac {4}{5}×\frac {6}{5}$;
……
(1)用你发现的规律填写下列式子的结果:
$1-\frac {1}{6^{2}}= $____×____;$1-\frac {1}{10^{2}}= $____×____.
(2)用你发现的规律计算:$(1-\frac {1}{2^{2}})×(1-\frac {1}{3^{2}})×(1-\frac {1}{4^{2}})×... ×(1-\frac {1}{2019^{2}})×(1-\frac {1}{2020^{2}})$.

答案：
(1)$\frac{5}{6}$ $\frac{7}{6}$ $\frac{9}{10}$ $\frac{11}{10}$
(2)解:原式$=\frac{1}{2}×\frac{3}{2}×\frac{2}{3}×\frac{4}{3}×\frac{3}{4}×\frac{5}{4}×\cdots×\frac{2018}{2019}×\frac{2020}{2019}×\frac{2019}{2020}×\frac{2021}{2020}=\frac{1}{2}×\frac{2021}{2020}=\frac{2021}{4040}$.