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Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)
Ta có : \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{16}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}=\frac{1}{10000}< \frac{1}{99.100}\)
Suy ra : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy ta có đpcm
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{100}{101}:2\)
\(A=\frac{50}{101}< 1\left(đpcm\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}\)
\(\Rightarrow B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow B< 1-\frac{1}{100}\)
\(\Rightarrow B< \frac{99}{100}< 1\left(đpcm\right)\)
a, M=1/1.2+1/2.3+...+1/49.50
M=1−1/2+1/2−1/3+...+1/49−1/50
M=1−1/50<1
Vậy M<1
\(a,\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}< 1\)
\(=>M< 1\)
câu a
Gọi ƯCLN (12n+1,30n+2) là d
⇒(12n+1)⋮d
(30n+2)⋮d
⇒5(12n+1)−2(30n+2)⋮d
⇒60n+5−60n−4⋮d
⇒1⋮d⇔d=1
Vậy ƯCLN (12n+1,30n+2)=1⇔12n+1/30n+2 là p/s tối giản
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}< 1\)
\(\Rightarrow A< 1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A< 1-\frac{1}{10}\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Vậy \(A< 1\)
mk cho tổng quát bn dựa vào mà lm nghen:
1/b bình < 1/b-1 - 1/b
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2S-S=1-\frac{1}{2^{100}}\)
\(\Rightarrow S=1-\frac{1}{2^{100}}< 1\) (đpcm)
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