trả lời giup mk

Ta có:
B=1-1/2²-1/3²-...-1/2004²
=1-(1/2²+1/3²+...+1/2004²)
=1-[1/(2.2)+1/(3.3)+...+1/(2004.2004)]
Ta thấy:
1/(2.2)>1/(2.3)
1/(3.3)>1/(3.4)
...
1/(2004.2004)>1/(2004.2005)
Cộng từng vế của các bất đẳng thức trên ta được:
1/(2.2)+1/(3.3)+...+1/(2004.2004) > 1/(2.3)+1/(3.4)+...+1/(2004.2005) = 1/(3.2)+1/(4.3)+...+1/(2005.2004)
= (3-2)/(3.2)+(4-3)/(4.3)+...+(2005-2004)/(2005.2004)
=3/(3.2)-2/(3.2)+4/(4.3)-3/(4.3)+...+2005/(2005.2004)-2004/(2005.2004)
=1/2-1/3+1/3-1/4+...+1/2004-1/2005
=1/2-1/2005
=2003/4010
=> B>1-2003/4010=2007/4010>2007/4022028=1/2004
Hay B>1/2004

Ta có :

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(2B=1-\frac{1}{3^{2005}}< 1\)

\(\Rightarrow\frac{2B}{2}=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)

\(\Rightarrow B< \frac{1}{2}\)

Đề: cmr: B = 1 - 1/2² - 1/3² - 1/4² -...-1/2004² > 1/2004 ( bn có ghi nhầm đề ko z)

Bài làm

ta có: \(\frac{1}{2^2}>\frac{1}{1.2};\frac{1}{3^2}>\frac{1}{2.3};\frac{1}{4^2}>\frac{1}{3.4};...;\frac{1}{2004^2}>\frac{1}{2003.2004}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)= 2003/2004

\(\Rightarrow B=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\frac{2003}{2004}=\frac{1}{2004}\)

=> đpcm

@I don't need you: Hey \(\frac{1}{2^2}>\frac{1}{1.2}\Leftrightarrow0.25>0.5?!?\)

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

          Giải

Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2004^2}< \frac{1}{2003.2004}\)

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

\(>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\right)\)

\(=1-\left(1-\frac{1}{2004}\right)=\frac{1}{2004}\) (đpcm)

xl mk làm sai r

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:

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

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

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

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).

\(\Rightarrowđpcm.\)

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

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

Đặt :

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

Ta có :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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