chứng minh rằng :( 1/2^3 + 1/3^3 +.....+ 1/2005^3 + 1/2006^3 )< 1/4
Những câu hỏi liên quan
Chứng minh rằng: \(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}...+\frac{1}{2005^3}+\frac{1}{2006^3}>\frac{1}{15}\)
Chứng minh rằng:
a) 1/n^3<1/(n-1)n(n+1) với n là số nguyên lớn hơn 1
b)1/2^3+1/3^3+1/4^3+...+1/2005^3+1/2006^3<1/4
giúp mk vs nha!!!
Cho A =1/1-1/2+1/3-1/4+...+1/2005+1/2006
B=(1/1+1/2+1/2006)-2x(1/2+1/4+...+1/2006)
a)So sánh A và B
b)Chứng minh rằng A=1/1004+1/1005+...1/2006
a,\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=B\left(ĐPCM\right)\)
b, \(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1003}\right)\)
\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
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ui ghi lộn, chữ đpcm chuyển xuống dòng cuối cùng nhé :v
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Chứng minh rằng :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}_{ }\)
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)
\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)
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Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006
=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)
=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)
=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)
=>A=1/1004+1/1005+.....+1/2006
Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )
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Chứng minh: A=\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2005^3}+\frac{1}{2006^3}<\frac{1}{4}\)
A=(1-1/2^2)+(1-1/3^2)+(1-1/4^2)+...+(1-1/2005^2)+(1-1/2006^2). Chứng minh rằng A<1/2
Giúp mình với , mai mình thi rồi. Thanks
chứng minh rằng
\(1< \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{3n+1}< 2\)
\(\frac{3}{5}< \frac{1}{2004}+\frac{2}{2005}+\frac{2}{2006}+...+\frac{1}{4006}< \frac{3}{4}\)
Cho A = 1/1 - 1/2 + 1/3 - ... + 1/2005 - 1/2006
Chứng minh A = 1/1004 + 1/1005 + ... + 1/2006
=>A= 1+1/2+1/3+...+1/2006-2*(1/2+1/4+....+1/2006)
=>A=1+1/2+1/3+....+1/2006-(1+1/2+.....+1/1003)
=> A=1/1004+1/1005+.....+1/2006
study well
k nha
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2008+2007/2+2006/3+2005/4+2005/5+........................3/2006+2/2007+1/2008
1/2+1/3+1/4+1/5+....................+1/2009
2008-1/2008=2007/2008
1/2-1/2009=2007/2009
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