Các câu hỏi tương tự
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{n^2+\left(n+1\right)^2}\le\frac{1}{2}\)
với mọi n thuộc N
CMR: Với mọi số tự nhiên n ta luôn có: \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{9}{20}\)
frac{1}{3}-frac{3}{5}+frac{5}{7}-frac{7}{9}+frac{9}{11}-frac{11}{13}+frac{11}{15}+frac{11}{13}-frac{9}{11}+frac{7}{9}-frac{5}{7}+frac{3}{5}-frac{1}{3}frac{-3}{4}.31frac{11}{23}-0,75.8frac{12}{28}left(2frac{1}{3}+3frac{1}{2}right):left(-4frac{1}{6}+3frac{1}{7}right)+7frac{1}{2}4frac{5}{9}:left(-frac{5}{7}right)+5frac{4}{9}:left(-frac{5}{7}right)frac{2}{3}-4left(frac{1}{2}+frac{3}{4}right)left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)....left(1-frac{1}{n+1}right)nin Z
Đọc tiếp
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)\(\frac{-3}{4}.31\frac{11}{23}-0,75.8\frac{12}{28}\)\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)\(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)\(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)\)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{n+1}\right)n\in Z\)
Với mọi số tự nhiên n > 2 . Chứng minh rằng \(\frac{1}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}\left[\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}\right]\)
Bài 1 ; \(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+4+.....+2010}\)
Bài 2 : CHỨNG MINH RẰNG: Với mọi số nguyên n>1 , ta có :
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+.....+\frac{1}{n^2+\left(n+1\right)^2}< \frac{9}{20}\)
Chứng minh rằng với mọi n \(\inℕ^∗\):
D = \(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}< 1\)
F = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
Tìm n biết:
\(\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)=\frac{2013}{2014}\)
Với \(n\in\)N*
\(\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)\left(\frac{2}{4.5}\right)........\left(\frac{2}{n\left(n+1\right)}-1\right)\left(n\in N\ne0,n\ge2\right)\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)