Question

cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}\).chứng minh rằng A<3/4

Answer 1

Ta có : 

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)

Hình như có gì đó sai sai :')

Answer 2

A+1/4=1/2+1/3²+......+1/2011²

A+1/4<1/2+1/2*3 +1/3*4 +....1/2010*2011

A+1/4<1-1/2011<1=3/4+1/4

A<1/4 (ĐPCM)

Answer 3

NHẦM BẠN ƠI     A<3/4

Answer 4

de dung rui ma