Question

[SPECTRE]

CMR :

a) N = 1/4^2 + 1/6^2 + 1/8^2 + ... + 1/(2n)^2 < 1/4 ( n thuộc N ; n lớn hơn hoặc bằng 2 )

b) P = 2!/3! + 2!/4! + 2!/5! + ... + 2!/n! < 1 ( n thuộc N ; n lớn hơn hoặc bằng 3 )

Answer 1

[Thanh Tùng DZ]

a) \(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

 \(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )

\(\Rightarrow N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.1=\frac{1}{4}\)

Vậy \(N< \frac{1}{4}\)

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

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

\(P< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n-1\right).n}\right)\)

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

Vậy \(P< 1\)

Answer 2

[Phùng Quang Thịnh]

P<1 nha bn k nha