\(\frac{8}{2n}=2\)
LL
Những câu hỏi liên quan
1\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2n}=\frac{2n-1}{2n}\)
Sao bạn gi được phân số vậy
A=\(\frac{2n-9}{n-4}\)
B=\(\frac{4n+6}{2n+1}\)
C=\(\frac{n-8}{2+n}\)
D=\(\frac{n-8}{2-n}\)
\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(1^{2n-1}=1\cdot2:8\)
\(1^{2n-1}=\frac{1}{4}\) ( vô lí vì \(1^{2n-1}=1\forall n\)
Vậy không có n thỏa mãn
\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4.\left(1^{2n-1}\right)}{8}=\frac{1}{8}\)
\(\Leftrightarrow1^{2n-1}=\frac{1}{4}\)
\(\Leftrightarrow1^{2n}=\frac{1}{4}\)
\(\Leftrightarrow1^n.1^2=\frac{1}{4}\)
\(\Leftrightarrow n=-4\)
Ta có: \(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
Vì \(1^{2n-1}=1\)\(\Rightarrow\)\(\frac{1^{2n-1}}{2}=\frac{1}{2}\)mà \(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(\Rightarrow\)\(S=\varnothing\)
\(CMR:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=\frac{\left(2n-1\right)}{2^n}\)
Bạn tham khảo cách làm ở đây: https://olm.vn/hoi-dap/question/528628.html
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\(CMR:\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{n}\right)\)(đpcm)
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Ta có:\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{4.4}+\frac{1}{4.9}+\frac{1}{4.16}+...+\frac{1}{4.n^2}\)
\(=\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)\)
\(Xét:\)
\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{n.n}< \frac{1}{\left(n-1\right).n}...\)
\(Suyra:\)
\(P=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(\Leftrightarrow P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Leftrightarrow P< 1-\frac{1}{n}< 1\)
\(\Leftrightarrow\frac{1}{4}.P< 1.\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)
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DH
20 tháng 3 2016 lúc 18:39
Tìm n để biểu thức sau là số nguyên :
\(A=\frac{2n+1}{n+2}-\frac{n+1}{n+2}+\frac{3n+5}{2n+4}+\frac{4n+6}{3n+6}-\frac{10n+12}{5n+10}-\frac{12n+3}{4n+8}\)
Chứng minh :\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\).... \(+\frac{1}{\left(2n\right)^2}\)= \(\frac{1}{2^2}\). ( \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}\)) < \(\frac{1}{2^2}\)( \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).\left(n\right)}\)) = \(\frac{1}{2^2}\)( \(1-\frac{1}{n}\)) < \(\frac{1}{2^2}\).1 = \(\frac{1}{4}\)
\(\Rightarrow\)\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{4}\)
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KN
16 tháng 3 2019 lúc 10:08
CMR: \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
5/ lim \(\frac{\left(12-n\right)^3\left(n-2\right)}{\sqrt{n^8-1}-2n^4}\)
6/ lim \(\frac{\sqrt[3]{3-8n^3}-n}{2n+5}\)
7/ lim \(\frac{\sqrt{n^6-2n+1}}{\sqrt{4n^6+3n}}\)
8/ lim \(\left(n^4+2n-20\right)\)
Chứng minh \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< 4\)