Question

cho

P=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+.....+\frac{2!}{n!}\)

         P<1(n\(\in\) N*)

Answer 1

\(P=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}=2!.\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...+\frac{1}{n!}\right)\)

\(< 2!.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n-1\right).n}\right)=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{n}\right)=1-\frac{1}{n}< 1\)

=>điều phải chứng minh