\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{1317}-\frac{1}{1318}+\frac{1}{1319}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{1319}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{1318}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{1319}-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{658}+\frac{1}{659}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{659}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{659}\right)+\frac{1}{660}+\frac{1}{661}+......+\frac{1}{1319}\)

\(=\frac{1}{660}+\frac{1}{661}+.........+\frac{1}{1319}\)

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 \(P< 1\)

P<1 nha bn k nha

Ta có :

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

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

\(M< 1-\frac{1}{n}\)

Mà \(1-\frac{1}{n}< 1\)nên M < 1

Vậy ...

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

........

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

\(\Rightarrow M=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}=\frac{n-1}{n}< 1\) (đpcm)

Vãi em mới học lớp 6 mà cô cho em bài này để ôn thi giữa kỳ

adasd