Question

[Duy Hoàng]

Cho \(n\in Z^+;n>1\)

Đặt \(P=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)....\left(1-\dfrac{1}{1+2+...+n}\right)\)

Tìm n để \(\dfrac{1}{P\left(n\right)}\in N\)

Answer 1

[Akai Haruma]

Lời giải:

Xét một thừa số tổng quát:

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

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

Do đó:

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

\(P_n=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{(n-1)(n+2)}{n(n+1)}\)

\(P_n=\frac{(1.2.3...(n-1))(4.5.6...(n+2))}{(2.3.4...n)(3.4.5..(n+1))}\)

\(P_n=\frac{1}{n}.\frac{n+2}{3}=\frac{n+2}{3n}\Rightarrow \frac{1}{P_n}=\frac{3n}{n+2}\)

Để \(\frac{1}{P_{n}}\in\mathbb{N}\Rightarrow \frac{3n}{n+2}\in\mathbb{N}\)

\(\Leftrightarrow 3n\vdots n+2\)

\(\Leftrightarrow 3(n+2)-6\vdots n+2\)

\(\Leftrightarrow 6\vdots n+2\)

\(\Rightarrow n+2=6\) do \(n+2>3\forall n>1\)

\(\Leftrightarrow n=4\)

Vậy \(n=4\)