đặt Pn= \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+....+n}\right)\)
Tìm tất cả các số nguyên dương n (n>1) sao cho \(\frac{1}{P_n}\)là số nguyên
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13 tháng 2 2018
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
2T
30 tháng 8 2019
Đặt \(a-b=x;b-c=y;c-a=z\)
\(\Rightarrow x+y+z=a-b+b-c+c-a=0\)
Lúc đó: \(B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Mà \(x+y+z=0\Rightarrow2\left(x+y+z\right)=0\Rightarrow\frac{2\left(x+y+z\right)}{xyz}=0\)
\(\Rightarrow B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{xz}+\frac{2}{xy}\)
\(=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
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!
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow P_n=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(P_n=\frac{1.2.3...\left(n-1\right)}{2.3.4...n}.\frac{4.5...\left(n+2\right)}{3.4...\left(n+1\right)}=\frac{n+2}{3n}\)
\(\Rightarrow\frac{1}{P_n}=\frac{3n}{n+2}=3-\frac{6}{n+2}\in Z\)
\(\Rightarrow n+2=Ư\left(6\right)=\left\{3;6\right\}\Rightarrow n=\left\{1;4\right\}\)