Question

Chứng minh rằng:

a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}<1\) 

b) \(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}>\frac{13}{21}\)

Answer 1

Bạn đổi phân số thành / rồi tìm trên Google có đầy bài này rồi.

Answer 2

a, VT < 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/2007.2008

          = 1-1/2+1/2-1/3+1/3-1/4+....+1/2007-1/2008 = 1-1/2008 < 1

=> ĐPCM

Answer 3

a) Ta có :1/2² + 1/3² + 1/4² + ... + 1/2008² < 1-1/2+1/2-1/3+...+1/2007-1/2008=1-1/2008<1

=> ĐPCM

Answer 4

\(a)\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}< 1\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2008.2008}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2007.2008}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2008.2008}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2007}-\frac{1}{2008}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2008.2008}< 1-\frac{1}{2008}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}< 1\)