Chứng minh\(\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{6}\right)\)\(\left(1-\frac{1}{10}\right)\)... \(\left(1-\frac{1}{253}\right)\)<\(\frac{2}{5}\)Mọi người giúp em bài này với ạ! Em cảm ơn!
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ML
29 tháng 3 2016
\(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\left(1-\frac{2}{2.3}\right)\left(...\right).....\left[1-\frac{2}{n\left(n+1\right)}\right]=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{\left(n-2\right)\left(n+1\right)}{\left(n-1\right).n}.\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}=\)
\(=\frac{1}{3}.\frac{n+2}{n}=\frac{1}{3}-\frac{1}{3}.\frac{2}{n}>\frac{1}{3}\)
S
20 tháng 1 2018
\(=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot\cdot\cdot\cdot\frac{-35}{36}\)
\(=\frac{-4}{6}\cdot\frac{-10}{12}\cdot\frac{-18}{20}\cdot\cdot\cdot\cdot\frac{-70}{72}\)
\(=\frac{-1.4}{2.3}\cdot\frac{-2.5}{3.4}\cdot\frac{-3.6}{4.5}\cdot\cdot\cdot\cdot\frac{-7.10}{8.9}\)
\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-7\right)}{2.3.4....8}\cdot\frac{4.5.6....10}{3.4.5....9}\)
\(=\frac{\left(-1\right).2.3...7}{2.3.4....8}\cdot\frac{10}{3}\)
\(=\frac{-1}{8}\cdot\frac{10}{3}=\frac{-5}{12}\)
13 tháng 6 2020
(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)
=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10
=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10
=1/10
LN
24 tháng 2 2019
\(B=\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).......\left(\frac{1}{1225}-1\right)\left(\frac{1}{1275}-1\right)\)
\(B=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}......\frac{-1224}{1225}.\frac{-1274}{1275}\)
\(B=\frac{-4}{6}.\frac{-10}{12}.\frac{-18}{20}......\frac{-2448}{2450}.\frac{-2548}{2550}\)
\(B=\frac{-4}{2.3}.\frac{-10}{3.4}.\frac{-18}{4.5}.....\frac{-2448}{49.50}.\frac{-2548}{50.51}\)
\(\Rightarrow\)B có : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )
\(\Rightarrow B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}........\frac{2448}{49.50}.\frac{2548}{50.51}.\left(-1\right)\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.........\frac{48.51}{49.50}.\frac{49.52}{50.51}.\left(-1\right)\)
\(B=\frac{\left(1.2.3...48.49\right).\left(4.5.6......51.52\right)}{\left(2.3.4......49.50\right).\left(3.4.5.....50.51\right)}.\left(-1\right)\)
\(B=\frac{52}{50.3}.\left(-1\right)\)
\(B=\frac{26}{75}.\left(-1\right)\)
Vậy \(B=\frac{-26}{75}\)