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7 tháng 7 2017

Đặt :

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

\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{98}}\)

\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{99}}\right)\)\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}< 1\)

\(\Leftrightarrow A< 1\)

Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{99}}< 1\rightarrowđpcm\)

7 tháng 7 2017

Đặt:

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

\(3S=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)

\(3S=1+\dfrac{1}{3}+.....+\dfrac{1}{3^{98}}\)

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

\(2S=1-\dfrac{1}{3^{99}}\)

\(2S< 1\)

\(S< 1\rightarrowđpcm\)

18 tháng 2 2022

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 

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:

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

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

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

11 tháng 2 2023

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

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

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

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

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

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

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

13 tháng 5 2018

Đáp án nè:

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

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

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

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

4A bé hơn(sorry tớ không thấy dấu bé hơn)\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

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

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

4B=\(3-\dfrac{1}{3^{99}}\) bé hơn 3 \(\Rightarrow\)B bé hơn \(\dfrac{3}{4}\)

\(\Rightarrow\) 4A bé hơn\(\dfrac{3}{4}\Rightarrow\)A bé hơn \(\dfrac{3}{16}\)

Tick cho mình nha , ngồi đánh máy tính mỏi cả mắt lun

Chúc học tốtvui

24 tháng 4 2017

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

11 tháng 5 2017

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>

11 tháng 5 2017

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 hihi!