tìm tổng :
-1-\(\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2004}\)
Những câu hỏi liên quan
Bài 1:
1) Tìm số nguyên dương a lớn nhất sao cho 2004! chia hết cho 7a.
2) Tính \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Chứng minh rằng tổng :
\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
1. Tính : P =\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{5}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Bài 2 : Tính : B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
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Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
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Tính:
B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+..+\frac{1}{2004}}\)
CMR :
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}\)
\(=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\right)\)
\(>1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\right)\)
\(>1-\left(1-\frac{1}{2004}\right)\)
\(>1-1+\frac{1}{2004}\)
\(>\frac{1}{2004}\left(đpcm\right)\)
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\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}}\). Tính giá trị bt trên
\(\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}=\left(2004-1-1-...-1\right)+\left(\frac{2003}{2}+1\right)+...+\left(\frac{1}{2004}+1\right)\)
\(=1+\frac{2005}{2}+...+\frac{2005}{2014}=2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2004}\right)\)
vậy \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}\right)}=\frac{1}{2005}\)
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CM rằng tổng
S=\(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
Ai nhanh và đúng mk sẽ tick 3 tick:))
CMR tổng
S=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{2^6}-...+\frac{1}{4n-2}-\frac{1}{4n}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0.2\)
Jup mk vs jk