CM: \(\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+..........+\frac{2n+1}{n^2.\left(n+1\right)^2}\)<1
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HT
1
AM
18 tháng 6 2015
\(G=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+....+\frac{2n+1}{n^2.\left(n+1\right)^2}=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{2n+1}{n^2\left(n^2+2n+1\right)}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{n^2}-\frac{1}{n^2+2n+1}\)
\(=1-\frac{1}{n^2+n+1}\left(n>0\right)\Rightarrow1-\frac{1}{n^2+n+1}
25 tháng 3 2018
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
30 tháng 11 2019
Minh AnNgọc HnueBăng Băng 2k6Thảo PHồ Đđề bài khó wáỖ CHÍ DŨNGBảo TrâmhLương Minh HằngươngAnh Qua
TG
30 tháng 11 2019
c/
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{-5}{3}:\frac{-10}{3}\)
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{-5}{3}.\frac{-3}{10}\)
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{1}{2}\)
\(=1-\left(\frac{66}{84}+\frac{98}{84}-\frac{70}{84}-\frac{42}{84}\right)\)
24 tháng 3 2016
\(\begin{equation} x = a_0 + \cfrac{1}{740_1 + \cfrac{1}{897654_2 + \cfrac{1}{672_3 + \cfrac{1}{100_4} } } } \end{equation}\)
AH
Akai Haruma
Giáo viên
10 tháng 4 2020
1.
\(\lim \frac{3n^2+5n+4}{2-n^2}=\lim \frac{\frac{3n^2+5n+4}{n^2}}{\frac{2-n^2}{n^2}}=\lim \frac{3+\frac{5}{n}+\frac{4}{n^2}}{\frac{2}{n^2}-1}=\frac{3}{-1}=-3\)
2.
\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)
3.
\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)
\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)
4.
\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)
5.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)
Ta thấy \(\frac{3}{4}=\frac{1}{1^2}-\frac{1}{2^2};\frac{5}{36}=\frac{1}{2^2}-\frac{1}{3^2};...\)
Tổng quát: \(\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
Đặt \(A=\frac{3}{4}+\frac{5}{36}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(\Rightarrow A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Do \(\left(n+1\right)^2>0\Rightarrow A< 1.\)