Pn=\(\frac{2}{3}\times\frac{5}{6}\times...\times\frac{\frac{\left(n+1\right)n}{2}-1}{\frac{\left(n+1\right)n}{2}}\)
= \(\frac{4}{6}\times\frac{10}{12}\times...\times\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
= \(\frac{1\times4}{2\times3}\times\frac{2\times5}{3\times4}\times...\times\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
= \(\frac{1\times2\times...\times\left(n-1\right)}{2\times3\times...\times n}\times\frac{4\times5\times...\times\left(n+2\right)}{3\times4\times...\times\left(n+1\right)}\)
= \(\frac{1}{n}\times\frac{n+2}{3}\)
=\(\frac{n+2}{3n}\)
=> \(\frac{1}{Pn}\)=\(\frac{3n}{n+2}\)
Đến đây thì bạn tự giải tiếp nhé.
Chúc bạn học tốt!
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow P_n=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(P_n=\frac{1.2.3...\left(n-1\right)}{2.3.4...n}.\frac{4.5...\left(n+2\right)}{3.4...\left(n+1\right)}=\frac{n+2}{3n}\)
\(\Rightarrow\frac{1}{P_n}=\frac{3n}{n+2}=3-\frac{6}{n+2}\in Z\)
\(\Rightarrow n+2=Ư\left(6\right)=\left\{3;6\right\}\Rightarrow n=\left\{1;4\right\}\)
