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*
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chịu thôi bạn ạ ko hiểu gì hết
\(\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...........\left(1+\frac{1}{n\left(n+2\right)}\right)=\frac{2013}{2014}\)
Tìm n , n thuộc N
cmr\(B=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left[1+\frac{1}{n+\left(n+2\right)}\right]< 2\)
Chứng minh rằng với mọi số tự nhiên n\(\ge\)1
Tính Q=\(\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+.....+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+......+\frac{1002.1004}{2005.2007}\)
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
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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 x, biết \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{4016}{2007}\)
