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Đặt S=1/4+1/16+1/36+...+1/10000
S= 1/4x(1+1/4+1/9+...+1/2500)
S= 1/4x(1+1/2x2+1/3x3+...+1/50x50)
S< 1/4x(1+1/1x2+1/2x3+...1/49x50)
S< 1/4x(1+1-1/2+1/2-1/3+....+1/49-1/50)
S< 1/4x(1+1-1/50)
S< 1/4x(2-1/50)<2/4(2/4=1/2)
S< 1/2
Ta có: \(\frac{1}{4}< \frac{1}{2}\)
\(\frac{1}{16}< \frac{1}{2}\)
... . . .
\(\frac{1}{10000}< \frac{1}{2}\)
\(\frac{1}{10000}+\frac{1}{10000}+...+\frac{1}{10000}< \frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\)(*) (n phân số \(\frac{1}{10000}\) ; n phân số \(\frac{1}{2}\))
Từ đó suy ra \(\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{1000}< \frac{1}{2}\left(đpcm\right)\)
Giải thích thêm: ta thấy \(\frac{1}{2^2}>\frac{1}{100}\),...,\(\frac{1}{10^2}=\frac{1}{100}\)=> từ \(\frac{1}{2^2}\)đến \(\frac{1}{10^2}\)có 5 cặp
\(\frac{1}{12^2}< \frac{1}{100}\),...,\(\frac{1}{100^2}< \frac{1}{100}\)=> từ \(\frac{1}{12^2}\)đến \(\frac{1}{100^2}\)có 45 cặp
=> 45>5 => tổng < 1/2 (kết hợp với cái kia nx thì bn mới hiểu)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{20}\)
\(\Rightarrow A< 1-\frac{1}{20}< 1\)
a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)
a/b= 7/6 + 7/10 + 7/12
a/b= 7(1/6+1/10+1/12)
Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< 1-\frac{1}{8}< 1\)
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
Ta có\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\)<\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\)=\(\frac{5}{6}\)(6 c/s \(\frac{1}{5}\))
Ta lại có \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{17}\)<\(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\)=\(\frac{7}{11}\)(7 c/s \(\frac{1}{11}\))
Suy ra \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)<\(\frac{110}{55}\)=2
Vậy...
Hok tốt
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< \frac{6}{5}+\frac{7}{11}\)
\(\Rightarrow A< \frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\)( ĐPCM )
cm \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}< \frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\right)< \frac{1}{2}\)
\(\Leftrightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}< 2\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}< 1\)
ta cần chứng minh điều trên:
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}=1-\frac{1}{5}\)
\(\Leftrightarrow A< 1-\frac{1}{5}< 1\)
suy ra đpcm
Đặt biểu thức là A
Ta có:A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}\)
\(\Rightarrow\)A=\(\frac{1}{4}+\left(\frac{1}{16}+\frac{1}{36}\right)+\left(\frac{1}{64}+\frac{1}{100}\right)\)
\(\Rightarrow\)A<\(\frac{1}{2}\)