K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

26 tháng 4 2017

ta có \(1-\dfrac{1}{2^2}-..-\dfrac{1}{2014^2}=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2014^2}\right)\)

\(\Rightarrow B< 1-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)

\(\Rightarrow B< 1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\)

\(\Rightarrow B< 1-\left(1-\dfrac{1}{2014}\right)=1-1+\dfrac{1}{2014}=\dfrac{1}{2014}\)

\(\Rightarrow B< \dfrac{1}{2014}\left(dpcm\right)\)

26 tháng 4 2017

bài này hay nà

27 tháng 4 2017

Sửa đề:

CMR: \(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)

Giải:

Ta có:

\(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)

\(=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}\right)\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2004^2}\)

Dễ thấy:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)

\(.............................\)

\(\dfrac{1}{2004^2}=\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)

Cộng các vế trên với nhau ta được:

\(A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2004.2005}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2004}-\dfrac{1}{2005}\)

\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2005}=2\)

27 tháng 4 2017

Chết!

\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2005}=\dfrac{2003}{4010}\)

Còn lại tự giải thôi! Dễ rồi

11 tháng 1 2018

Ta có:

\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-........-\dfrac{1}{2004^2}.\)

\(B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}\right).\)

Đặt \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}.\)

Ta thấy:

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}.\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}.\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}.\)

..................

\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}.\)

\(\Rightarrow M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}.\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+........+\dfrac{1}{2003.2004}.\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+........+\dfrac{1}{2003}-\dfrac{1}{2004}.\)

\(=\dfrac{1}{1}-\dfrac{1}{2004}.\)

\(=\dfrac{2003}{2004}.\)

\(\Rightarrow M< \dfrac{2003}{2004}.\)

\(\Rightarrow1-M>1-\dfrac{2003}{2004}.\)

\(\Rightarrow B>\dfrac{1}{2004}\) (do B = 1 - M).

\(\Rightarrowđpcm.\)

11 tháng 1 2018

\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)

\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...........+\dfrac{1}{2004^2}\right)\)

Đặt :

\(H=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{2004^2}\)

Ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

.......................

\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+........+\dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)

\(\Leftrightarrow A< \dfrac{2003}{2004}\)

\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)

\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)

AH
Akai Haruma
Giáo viên
28 tháng 4 2018

Lời giải:

Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)

\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)

\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)

--------------------------

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)

\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)

\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)

Vậy ta có đpcm.

19 tháng 4 2018

Tao có: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)

\(B>1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\right)\)

\(B>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)

\(B>1-\left(1-\dfrac{1}{2004}\right)=1-1+\dfrac{1}{2004}=\dfrac{1}{2004}\left(đpcm\right)\)

25 tháng 4 2023

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

25 tháng 4 2023

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2

9 tháng 7 2017

a)

\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)

\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..........................

\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)

\(\Leftrightarrow A< \dfrac{2003}{2004}\)

\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)

\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)

b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)

\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)

\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)

\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)

\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)

\(\Leftrightarrow S< 0,2\left(đpcm\right)\)

19 tháng 2 2020

cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik

2 tháng 5 2021

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...............

\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

 

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Lời giải:
$B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}$

$B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{8-7}{7.8}$

$B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$

$B< 1-\frac{1}{8}$

Mà $1-\frac{1}{8}< 1$ nên $B< 1$ (đpcm)

30 tháng 4 2023

 

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{8^2}< 1\)

\(\dfrac{1}{2^2}< \dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

.......

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

\(=>B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{8^2}< \dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.....+\dfrac{1}{7\cdot8}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+.....+\dfrac{1}{7}-\dfrac{1}{8}=1-\dfrac{1}{8}< 1\)

\(=>B< 1\)