\(1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
\(=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(P=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\cdot...\cdot\left(1-\dfrac{1}{1+2+...+n}\right)\)
\(=\left(1-\dfrac{1}{2\cdot\dfrac{3}{2}}\right)\left(1-\dfrac{1}{3\cdot\dfrac{4}{2}}\right)\cdot...\cdot\left(1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}\right)\)
\(=\left(1-\dfrac{2}{2\cdot3}\right)\left(1-\dfrac{2}{3\cdot4}\right)\cdot...\cdot\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(=\dfrac{\left(2+2\right)\left(2-1\right)}{2\cdot\left(2+1\right)}\cdot\dfrac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot...\cdot\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(=\dfrac{4\cdot1}{2\cdot3}\cdot\dfrac{5\cdot2}{3\cdot4}\cdot...\cdot\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(=\dfrac{4\cdot5\cdot...\cdot\left(n+2\right)}{3\cdot4\cdot...\cdot\left(n+1\right)}\cdot\dfrac{1\cdot2\cdot...\cdot\left(n-1\right)}{2\cdot3\cdot...\cdot n}\)
\(=\dfrac{n+2}{3}\cdot\dfrac{1}{n}=\dfrac{n+2}{3n}\)
Đặt Q=1:P
\(=1:\dfrac{n+2}{3n}=\dfrac{3n}{n+2}\)
Để Q nguyên thì \(3n⋮n+2\)
=>\(3n+6-6⋮n+2\)
=>\(-6⋮n+2\)
mà n+2>=4(n>=2)
nên \(n+2=6\)
=>n=4
Ta thấy :
\(1-\dfrac{1}{1+2+...+n}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(P=\dfrac{1}{2}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow P=\dfrac{1}{2}.\dfrac{n+2}{n}=\dfrac{n+2}{2n}\)
\(\Rightarrow\dfrac{1}{P}=\dfrac{2n}{n+2}\in Z\)
\(\Leftrightarrow2n⋮n+2\)
\(\Leftrightarrow2n-2n-4⋮n+2\)
\(\Leftrightarrow-4⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;2;4\right\}\)
\(\Leftrightarrow n\in\left\{-1;0;2\right\}\Rightarrow n=2\left(n\in Z^+\right)\)
Vậy \(n=2\) thỏa đề bài
