\(\frac{1}{11^2}+\frac{1}{12^2}+\frac{1}{13^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)

\(=\frac{1}{11.11}+\frac{1}{12.12}+\frac{1}{13.13}+\frac{1}{14.14}+...+\frac{1}{100.100}\)

\(< \frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{99.100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}\)

Vì \(\frac{1}{100}>0\Rightarrow\frac{1}{10}-\frac{1}{100}< \frac{1}{10}\)

\(\RightarrowĐPCM\)

theo mình tình thi  \(\frac{1}{11^2}+\frac{1}{12^2}+......+\frac{1}{100^2}=0,08521616902\)

mà \(\frac{1}{10}=0,1\)

\(\Rightarrow0,08521515902< 0,1\)

sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)

\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)

Đề gõ sai, xin sửa lại:
Chứng minh:

\({1 \over {11}^2} + {1 \over {12}^2} + {1 \over {13}^2} + {1 \over {14}^2} + ... + {1 \over {100}^2}<{1 \over {10}}\)

Cảm ơn

Đặt biểu thức là A     ta có:

1/11^2 < 1/10.11 = 1/10 - 1/11

1/12^2 < 1/11.12 = 1/11 - 1/12

1/13^2 < 1/12.13 = 1/12 - 1/13

. . . . . . . . . 

1/100^2 < 1/99.100 = 1/99 - 1/100

 => A < 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + . . . .+ 1/99 - 1/100

  => A < 1/10 -  1/100

 => A < 1/10

                Bạn nhớ k cho mình nha

Có\(\frac{1}{11^2}\)<\(\frac{1}{10.11}\);...;\(\frac{1}{100}\)<\(\frac{1}{99.100}\)\(\Rightarrow\)\(\frac{1}{11^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)+.......+\(\frac{1}{99}\)-\(\frac{1}{100}\)<\(\frac{1}{10}\)-\(\frac{1}{100}\)

có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)

\(A< \frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{100.101}\)

\(A< \frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{101}\)

\(A< \frac{1}{10}-\frac{1}{101}=\frac{101}{1010}-\frac{10}{1010}=\frac{91}{1010}< \frac{505}{1010}\)

\(A< \frac{1}{2}\)

a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm

b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}