chung minh : 1/5^3+1/6^3+1/7^3+.........+1/2004^3 < 1/40
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chung minh : 1/5^3+1/6^3+1/7^3+........+1/2004^3 <1/40
Bài toán tổng quát:
Với mọi n\(\in\)N* ta có: \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{n\left(n^2-1\right)}=\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Áp dụng vào bài toán:
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}< \frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
mà \(\frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
\(=\frac{1}{2}\left(\frac{2}{4.5.6}+\frac{2}{5.6.7}+\frac{2}{6.7.8}...+\frac{2}{2003.2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+\frac{1}{6.7}-\frac{1}{7.8}...+\frac{1}{2003.2004}-\frac{1}{2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{2003.2004}\right)=\frac{1}{40}-\frac{1}{2.2003.2004}< \frac{1}{40}\)
=>\(\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}+...+\frac{1}{2002.2003.2004}< \frac{1}{40}\)
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chung minh rang 1/5^3+1/6^3+1/7^3+..........+1/2004^3<1/40
Chứng minh rằng : 1/65 < 1/5^3 + 1/6^3 + 1/7^3 + ... + 1/2004^3 <1/40
chứng minh rằng
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)
CMR:
\(\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{40}\)
Ta có: \(n^3-n< n^3\forall n\)
mà: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Nên: \(\left(n-1\right)n\left(n+1\right)< n^3\Leftrightarrow\dfrac{1}{\left(n-1\right)n\left(n+1\right)}>\dfrac{1}{n^3}\)
Trở lại bài toán:
\(SV=\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{\left(5-1\right).5.\left(5+1\right)}+\dfrac{1}{\left(6-1\right).6.\left(6+1\right)}+\dfrac{1}{\left(7-1\right).7.\left(7+1\right)}+...+\dfrac{1}{\left(2004-1\right).2004.\left(2004+1\right)}\)
\(SV< \dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+...+\dfrac{1}{2003.2004.2005}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}+\dfrac{1}{5.6}-\dfrac{1}{6.7}+\dfrac{1}{6.7}-\dfrac{1}{7.8}+...+\dfrac{1}{2003.2004}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2.4.5}-\dfrac{1}{2.2004.2005}=\dfrac{1}{40}-\dfrac{1}{2.2004.2005}< \dfrac{1}{40}\left(đpcm\right)\)
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1/5^3+1/6^3+...+1/2004^3<1/40
a)chung minh A= 2^1+2^2+2^3+2^4+...2^2010chia het cho 3
b)chung minh B= 3^1+3^2+3^3+3^4+...3^2010chia het cho 4
c)chung minh C= 5^1+5^2+5^3+5^4+...5^2010chia het cho 6
d)chung minh D= 7^1+7^2+7^3+7^4+...7^2010chia het cho 8
a) A=21+22+23+...+22010
A=(21+22)+(23+24)+.....+(22009+22010)
A=(21x3)+(23x3)+.....+(22009x3)
A=3x(21+23+.......+22009)
Vậy A chia hết cho 3.
NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !
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Chứng minh rằng:
\(\frac{1}{65}\)<\(\frac{1}{5^3}\)+\(\frac{1}{6^3}\)+\(\frac{1}{7^3}\)+...+\(\frac{1}{2004^3}\)<\(\frac{1}{40}\)
chứng minh rằng \(\frac{1}{65}\)<\(\frac{1}{5^3}+\frac{1}{6^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)