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Chứng tỏ rằng :\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
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)
4S=\(\dfrac{4}{2^2}-\dfrac{4}{2^4}+\dfrac{4}{2^6}-...+\dfrac{4}{2^{4n-2}}-\dfrac{4}{2^{4n}}+...+\dfrac{4}{2^{2002}}-\dfrac{4}{2^{2004}}\)
4S=1-\(\dfrac{1}{2^2}+\dfrac{1}{2^4}-,...-\dfrac{1}{2^{2002}}\)
4S+S=1-\(\dfrac{1}{2^{2004}}\)
5S=\(\dfrac{2^{2004}-1}{2^{2004}}\)<1
\(\Rightarrow\)5S<1 hay S<\(\dfrac{1}{5}\)=0,2(đpcm)
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005
B=1/2-1/(32005.2)
Vậy B <1/2
CHTT nha
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