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25 tháng 7 2021

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

 
20 tháng 4 2021

Đóng góp j vậy ạ

NV
21 tháng 4 2021

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

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:

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

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

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

NV
1 tháng 3 2023

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

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

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

\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\) (1)

\(\Rightarrow12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\) (2)

Cộng vế (1) và (2):

\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow16A< 3\)

\(\Rightarrow A< \dfrac{3}{16}\)

2 tháng 3 2023

Đặt `A` `=` `1/3 - 2/3^2+3/3^3 - 4/3^4+ ... + 99/3^99-100/3^100`
`=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99`
`=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100`
`=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99`
`=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...`
`=>16A=3-101/3^99-100/3^100`
`<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16`
`=> A<3/16`

@Nae

6 tháng 5 2017

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

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

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

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

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

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

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

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

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

6 tháng 5 2017

Đặt :

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

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

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

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

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

Đặt :

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

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

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

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

\(\Rightarrow4A< \dfrac{3}{4}\Rightarrow A< \dfrac{3}{16}\rightarrowđpcm\)

AH
Akai Haruma
Giáo viên
25 tháng 1 2018

Lời giải:

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

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

\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)

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

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

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

\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)

\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)

\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)

\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)