\(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(\left(1-\frac{2}{2.3}\right)\left(...\right).....\left[1-\frac{2}{n\left(n+1\right)}\right]=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{\left(n-2\right)\left(n+1\right)}{\left(n-1\right).n}.\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}=\)

\(=\frac{1}{3}.\frac{n+2}{n}=\frac{1}{3}-\frac{1}{3}.\frac{2}{n}>\frac{1}{3}\)

giúp mik với các bn ơi

CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm

Cảm ơn bạn

a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :

\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)

\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)

b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :

\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)

\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)