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Ta có:\(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4};.....;\frac{1}{100^2}< \frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\left(đpcm\right)\)
Gọi \(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< \frac{3}{4}\)
\(\Rightarrow D< \frac{3}{4}\left(đpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
Gọi ƯC 12n + 1 ; 30n + 2 là d
12n+1 chia hết cho d
30n + 2 chia hết cho d
=> (30n+2) chia hết cho d
=> 15n+1 chia hết cho d
<=> (15n+1) - (12n+1) chia hết cho d
<=> n thuộc ước của 3
n = -1 ; -3 ; 1 ; 3
p/s : chứng minh thô...
b) \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{13\cdot15}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(1-\frac{1}{15}\right)< \frac{1}{2}\)
\(\)
a, \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}< 1\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(........\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{n}\)\(=1-\frac{1}{n}=\frac{n-1}{n}< 1\)
a)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...-\frac{1}{2^6}=A\)
2A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}\)
2A + A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^6}\)
3A = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)(đpcm)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{100^2}< \frac{1}{1}+\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\\ \)
\(\frac{1}{1}+\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\\ =\frac{1}{1}+\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\\ =\frac{1}{1}+\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\\ =1\frac{3}{4}-\frac{1}{100}< 1\frac{3}{4}\)
Vậy \(A< 1\frac{3}{4}\)
Ta có với mọi n là số tự nhiên thì : \(\frac{1}{n^2}< \frac{1}{n\left(n+1\right)}\)
Áp dụng : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}< 1< 1\frac{3}{4}\)