Các câu hỏi tương tự
\(\sin^3\frac{x}{3}+3\sin^3\frac{x}{3^2}+...+3^{n-1}\sin^3\frac{x}{3}=\frac{1}{4}\left(3^n\sin^3\frac{x}{3^n}-\sin x\right)\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}\left(n\ge1\right)\)\(\left(n!\right)^2\ge n^2\ge\left(n+1\right)^{n-1}cho\left(n\ge1\right)\)
HB
22 tháng 9 2020 lúc 21:36
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...........\frac{2n-1}{2n}\)\(n\in N,n\ge2\)
C/m A<\(\frac{1}{\sqrt{3n+1}}\)
tính giới hạn sau : lim \(\frac{2n^3-n+3}{-3n^3-n+1}\)
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= \(\frac{m}{n^2}.\left(n^2-1\right):\frac{2mn}{n^2+1}\)
B= \(m:\frac{2mn^3-6mn^2+4mn}{n^4-3n^3+3n^2-3n+2}\)
Tính A+B
\(lim\frac{n^2-2n+3-\frac{1}{2^{n-1}}}{n^2}\)
Cho \(A=\frac{m}{m+1}.|n^2-1|.\frac{2mn}{n^2+1}\)
\(B=m:\frac{2mn^3-6mn^2+4mn}{n^4-3n^3+3n^2-3n+2}\)
Tính A + B
Chứng minh rằng: \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{\left(2n-1\right)}{2n}\le\frac{1}{\sqrt{3n+1}}\) ( n là số nguyên dương)
1) \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{\left(2n-1\right)}{2n}\le\frac{1}{\sqrt{3n+1}}\)( n là số nguyên dương)