\(CM:A< \frac{1}{6}với\)\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
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26 tháng 6 2020
\(A=\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)
\(\Leftrightarrow5A=\frac{1}{5}+\frac{2}{5^2}+......+\frac{99}{5^{99}}\)
\(\Leftrightarrow5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+....+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)
\(\Leftrightarrow4A=\frac{1}{5}+\frac{1}{5^2}+......+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt : \(H=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}\)
\(\Leftrightarrow5H=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\)
\(\Leftrightarrow5H-H=\left(1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)
\(\Leftrightarrow4H=1-\frac{1}{5^{99}}\)
\(\Leftrightarrow H=\frac{1}{4}-\frac{1}{4.5^{99}}< \frac{1}{4}\)
\(\Leftrightarrow4A< B< \frac{1}{4}\)
\(\Leftrightarrow A< \frac{1}{16}\left(đpcm\right)\)
10 tháng 6 2020
c) \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{2}.\frac{4}{4}.\frac{6}{6}...\frac{100}{100}=\frac{1}{2}\)
LH
13 tháng 5 2020
a) \(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)
\(4A=5A-A=\frac{1}{5}-\frac{1}{5^{2019}}\)
\(A=\frac{1}{20}-\frac{1}{4.5^{2019}}< \frac{1}{20}< \frac{1}{2}\)
b) Đề có sai không mà đằng cuối lại là \(\frac{1}{4^2}\)lặp lại lần nữa.
c) \(C=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(3C=2C+C=1-\frac{1}{64}< 1\)
\(C< \frac{1}{3}\)
d) Xem lại đề nữa đi e, nếu trừ hai vế cho \(\frac{1}{3}\)thì vế trái > 0 > vế phải rồi
e) \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)(10 số hạng)
\(=\frac{10}{50}=\frac{1}{5}\)
Tương tự: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{7}\)
\(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}>\frac{1}{8}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)
D
16 tháng 4 2015
b.Đặt A = \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{100^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)= \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)= \(\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}<\frac{25}{100}=\frac{1}{4}\)(1)
A > \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)(2)
Từ (1) và (2) =>\(\frac{1}{6}\) < A < \(\frac{1}{4}\)
mình cần gấp nhé, xin cảm ơn
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
<=> \(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
=> \(5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^5}+...+\frac{99}{5^{100}}\right)\)
=> \(4A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=> \(20A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
=> \(20A-4A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)
=> \(16A=1+\frac{99}{5^{100}}-\frac{100}{5^{99}}< 1\)
=> \(A< \frac{1}{16}< \frac{1}{6}\)