sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

Cộng vế với vế 

\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4 

Vậy ta có đpcm 

Đóng góp j vậy ạ

\(2^2< 2.3\Rightarrow\dfrac{1}{2^2}>\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

Tương tự: \(\dfrac{1}{3^2}>\dfrac{1}{3}-\dfrac{1}{4}\) ; \(\dfrac{1}{4^2}>\dfrac{1}{4}-\dfrac{1}{5}\) ; ....; \(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)

Do đó:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{101}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)

A= \(\dfrac{1}{3}-\dfrac{2}{3^2}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

3A= 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)

A + 3A = 1- \(\dfrac{1}{3}+\dfrac{1}{3^2}\) - \(\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

=> 4A < 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}\) \(\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

Đặt : B = 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

3B = 3 - 1 + \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}+.....+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)

B + 3B = 3 - \(\dfrac{1}{3^{99}}\)

4B = 3 - \(\dfrac{1}{3^{99}}\) < 3 => B < \(\dfrac{3}{4}\)

=> 4A < \(\dfrac{3}{4}\) => A < \(\dfrac{3}{16}\) ĐPCM

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(\Rightarrow A< 1.\left(\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}\right)\)

Mà ta thấy \(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow A< 1+\dfrac{3}{4}=\dfrac{7}{4}\)

Ta có

B = \(\dfrac{1}{2!}\) + \(\dfrac{2}{3!}\) + \(\dfrac{3}{4!}\) + ..... + \(\dfrac{99}{100!}\)

B = \(\dfrac{2-1}{2!}\) + \(\dfrac{3-1}{3!}\) + \(\dfrac{4-1}{4!}\) + ... + \(\dfrac{100-1}{100!}\)

B = \(\dfrac{2}{2!}\) - \(\dfrac{1}{2!}\) + \(\dfrac{3}{3!}\) - \(\dfrac{1}{3!}\) + ... + \(\dfrac{100}{100!}\) - \(\dfrac{1}{100!}\)

B = 1 - \(\dfrac{1}{2!}\) + \(\dfrac{1}{2!}\) - \(\dfrac{1}{3!}\) + ... + \(\dfrac{1}{99!}\)- \(\dfrac{1}{100!}\)

B = 1 - \(\dfrac{1}{100!}\) < 1

=> B < 1 <đpcm>

B=\(\dfrac{1}{2!}\)+\(\dfrac{2}{3!}+\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\)

=\(\dfrac{2-1}{2!}\)+\(\dfrac{3-1}{3!}+\dfrac{4-1}{4!}\)+...+\(\dfrac{100-1}{100!}\)

=\(\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+\dfrac{4}{4!}-\dfrac{1}{4!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)

=\(\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)

=\(1-\dfrac{1}{100!}\)< 1

\(\Rightarrow\)B =\(\dfrac{1}{2!}\)+\(\dfrac{2}{3!}+\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\) < 1

Chúc bạn học tốt !

Ta có :

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{2}{5}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.................+\dfrac{99}{100}}\)

\(=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+.............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+............+1-\dfrac{1}{100}}\)

\(=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+...........+\dfrac{2}{100}\right)}{\left(1+1+.........+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{100}\right)}\)

\(=\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..........+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.........+\dfrac{1}{100}\right)}\)

\(=2\)

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+..........+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}=2\rightarrowđpcm\)

đpcm là j ak