\(ChoA=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+......+\frac{1}{1+3+......+2013}\)
\(Cm:a<\frac{3}{4}\)
Những câu hỏi liên quan
Ai giúp mk nha:
\(ChoA=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2013}\)
CMR Nó <\(\frac{3}{4}\)
Thực hiện phép tính :
a ) A =\(\frac{2015+2013+2011+2009+...+7+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
b) B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
\(choA=\frac{2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}}.hỏiAchia3dưbaonhiêu\)
A=\(\frac{1+\frac{2011}{2}+1+\frac{2010}{3}+1+...+\frac{1}{2012}+1+1}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{2013\left(\frac{1}{2}+...+\frac{1}{2013}\right)}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=2013
Mà 2013: 3 = 671
Vậy A : 3 dư 0 hay\(A⋮3\)
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CMR:\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2013^3}< \frac{1}{40}\)
\(ChoA=\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2013}\) và B=\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)Tính\(\frac{A}{B}\)
Cho A=\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+......+\frac{1}{1+3+.....+2013}\). CMR A<\(\frac{3}{4}\)
Chứng minh rằng: \(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2013^3}< \frac{1}{40}\)
Cho A =\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+...+2013}\)
Chứng minh A < \(\frac{3}{4}\)
Cho \(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2013}\)
Cm nó <\(\frac{3}{4}\)Nha mn giúp mk nha còn có mỗi bài này nữa