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1/5^2 + 1/6^2 + 1/7^2 + ... + 1/100^2
< 1/4×5 + 1/5×6 + 1/6×7 + ... + 1/99×100
< 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/99 - 1/100
< 1/4 - 1/100 < 1/4 ( đpcm)
ok, ta co \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{4}-\frac{1}{100}\)
\(A< \frac{1}{4}\)
Lai co \(A>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}=\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}\)
\(A>\frac{1}{6}\)
Ta có\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}\)(A là đề bài)
Mà \(\frac{1}{5}-\frac{1}{30}=\frac{1}{6}< \frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{6}< A< \frac{1}{4}\left(ĐPCM\right)\)
Ta có: \(\frac{1}{5\cdot6}< \frac{1}{5^2}=\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)
\(\frac{1}{6\cdot7}< \frac{1}{6^2}=\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)
\(\frac{1}{7\cdot8}< \frac{1}{7^2}=\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)
.............................
\(\frac{1}{100\cdot101}< \frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
Đặt \(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(B=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(=>\frac{1}{6}< A< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< B< \frac{1}{4}\)
\(=>\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(Đpcm\right)\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)
Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)
Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)
Từ (1) và (2) =>đpcm
Đặt \(A=\displaystyle\sum_{i=5}^{100}\frac{1}{i^2}\)
\(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
sao cái mã latex ko hiển thị nhờ :(( A là cái biểu thức ở giữa nhé
\(Ta\) \(có : \)
\(1 / 5^2 + 1 /6^2 + ... + 1 /100^2 < 1 /4.5\)\(+ 1 / 5 .6 + ... + 1 / 99 .100\)
\(Mà ta có:\)\(1 / 4 .5 + 1 / 5 .6 + ... + 1 / 99 .100\)
\(\Rightarrow\)\(1 / 4 - 1 / 5 + 1 / 5 - 1 / 6 + ... +\)\(1 / 99 - 1 / 100\)
\(\Rightarrow\)\(1 / 4 - 1 / 100\) \(< 1 / 4\)
\(Nên 1 / 5^2 + 1 /6^2 + ...+ 1 / 100^2 < 1 / 4\)
Tương tự chứng minh tiếp nhé 😘😘
đặt \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=A\)
*chứng minh A<1/4
ta có:\(A<\frac{1}{4.5}+\frac{1}{5.6}+..+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\) *chứng minh A>1/6
ta có:
\(A>\frac{1}{5.6}+\frac{1}{6.7}+..+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
từ 2 điều trên =>đpcm
mk chắc chắn đúng,hồi chiều cô mk ms cho làm
đặt 1/5^2+1/6^2+...+1/100^2=A
ta có: \(A<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\left(1\right)\)
\(A>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+..+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\left(do\frac{1}{5}>\frac{1}{6}\right)\left(2\right)\)
từ (1);(2)=>1/6<A<1/4
=>đpcm