\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
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1 tháng 3 2020
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2A=1-\frac{1}{2n+1}\)
\(A=\frac{1}{2}-\frac{1}{\left(2n+1\right).2}< \frac{1}{2}\)
Vậy:...
- Hok tốt ~
I
1 tháng 3 2020
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}+\frac{1}{2n+1}\)
=>\(2A=1-\frac{1}{2n-1}\)
=>\(2A=\frac{2n}{2n+1}\)
=>\(A=\frac{2n}{4n+2}=\frac{2n}{2\left(n+1\right)}=\frac{n}{n+1}< \frac{1}{2}\)
zậy A<1/2
AH
Akai Haruma
Giáo viên
25 tháng 6 2019
Lời giải:
\(A=\frac{1}{1(2n-1)}+\frac{1}{3(2n-3)}+...+\frac{1}{(2n-3).3}+\frac{1}{(2n-1).1}\)
\(2nA=\frac{1+(2n-1)}{1(2n-1)}+\frac{3+(2n-3)}{3(2n-3)}+....+\frac{(2n-3)+3}{(2n-3).3}+\frac{(2n-1)+1}{(2n-1).1}\)
\(2nA=\frac{1}{2n-1}+1+\frac{1}{2n-3}+\frac{1}{3}+...+\frac{1}{3}+\frac{1}{2n-3}+1+\frac{1}{2n-1}\)
\(=\left(\frac{1}{2n-1}+\frac{1}{2n-3}+...+\frac{1}{3}+1\right)+\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)
\(=2\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
\(\Rightarrow A=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
YN
6 tháng 4 2019
\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)
Hok tốt
14 tháng 9 2017
a) \(\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\left(\frac{4}{8}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\frac{5}{8}\)
= \(\frac{8}{24}-\frac{15}{24}\)
= \(\frac{-7}{24}\)
b) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)
= \(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)\)+ \(\frac{1}{13}\)
= \(\left(\frac{4}{8}-\frac{2}{8}+\frac{1}{8}\right)+\frac{1}{13}\)
= \(\frac{1}{8}+\frac{1}{13}\)
= \(\frac{13}{104}+\frac{8}{104}\)
= \(\frac{23}{104}\)
c) \(13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)
= \(\left(13\frac{2}{7}+2\frac{5}{7}\right):\left(\frac{-8}{9}\right)\)
= \(16:\left(\frac{-8}{9}\right)\)
= -18
24 tháng 8 2019
có liền luôn nè
1/2 -43/101+(-1/3)-1/6
= -43/101+(-1/3)-1/6+1/2
=-43/101+0
=-43/101
PN
11 tháng 11 2015
Công thức tống quát:
\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)
Theo đó, ta có:
\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)
\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)
\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)
\(....................\)
\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)
Nhân lần lượt các đẳng thức trên, ta được:
\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)
1 tháng 4 2019
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2n-2\right).2n}\)
\(< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(< \frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
\(\Rightarrow\) \(A< \frac{1}{4}\)
Study well ! >_<
\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(2n=3+1\)
\(2n=4\)
\(\Rightarrow n=2\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
2n-1=3
2n=4
n=2