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

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(A\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

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

A<\(1-\dfrac{1}{100}=\dfrac{99}{100}\)(đpcm)

Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2.3},\dfrac{1}{3^2}>\dfrac{1}{3.4},...,\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

A>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)

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

A>\(\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)(đpcm)

Vậy \(\dfrac{99}{100}>A>\dfrac{99}{202}\)

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

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

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

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

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 

a: Đặt \(A=\frac12-\frac14+\frac18-\frac{1}{16}+\cdots-\frac{1}{1024}\)

=>\(A=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(2A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}\)

=>\(2A+A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}+\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(3A=1-\frac{1}{2^{10}}<1\)

=>\(A<\frac13\)

b: Đặt \(B=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>\(3B+B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>\(3A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>\(3A+A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

=>\(4A=-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}=\frac34-\frac{403}{4\cdot3^{100}}\)

=>\(4B<\frac34\)

=>\(B<\frac{3}{16}\)