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Ta có : \(\frac{5.5}{1.6}+\frac{5.5}{6.11}+\frac{5.5}{11.16}+\frac{5.5}{16.21}\)
\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)\)
\(=5\left(1-\frac{1}{21}\right)\)
\(=5.\frac{20}{21}=\frac{100}{21}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(>\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{51}\right)>\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{9}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{4}.\frac{3}{2}=\frac{3}{8}\)
\(\Rightarrow A>\frac{3}{8}\left(đpcm\right)\)
a) Ta có:
\(\frac{1}{n-1}-\frac{1}{n}=\frac{n-\left(n-1\right)}{n\left(n-1\right)}=\frac{1}{n\left(n-1\right)}>\frac{1}{n.n}=\frac{1}{n^2}\left(1\right)\)
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}< \frac{1}{n.n}=\frac{1}{n^2}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\frac{1}{n\left(n-1\right)}>\frac{1}{n^2}>\frac{1}{n\left(n+1\right)}\)
Hay \(\frac{1}{n-1}-\frac{1}{n}>\frac{1}{n^2}>\frac{1}{n}-\frac{1}{n+1}\) (Đpcm)
Ta có :
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{30}{40}=\frac{3}{4}\)
\(\Rightarrow\)\(A>\frac{3}{4}\) ( điều phải chứng minh )
Vậy \(A>\frac{3}{4}\)
Chúc bạn học tốt ~
A=\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
A=\(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{9}+\frac{1}{10}-\frac{1}{10}\)
A= 0
=> A>\(\frac{65}{132}\)
A = 1/5×5 + 1/6×6 + ... + 1/100×100
A < 1/4×5 + 1/5×6 + ... + 1/99×100
A < 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100
A < 1/4 - 1/100 < 1/4 (1)
A = 1/5×5 + 1/6×6 + ... + 1/100×100
A > 1/5×6 + 1/6×7 + ... + 1/100×101
A > 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/100 - 1/101
A > 1/5 - 1/101 > 1/5 - 1/30
A > 6/30 - 1/30 = 1/6 (2)
Từ (1) và (2) => 1/6 < A < 1/4 ( đpcm)