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Akai Haruma
Giáo viên
9 tháng 1
Lời giải:
\(F=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2010}{2009}.\frac{2011}{2010}\\ =\frac{3.4.5...2010.2011}{2.3.4...2009.2010}=\frac{2011}{2}\)
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
TD
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TM
0
18 tháng 2 2022
sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Cộng vế với vế
\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4
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