Question

Chứng minh rằng : 

a, \(\text{A}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{2}{4^{\text{2}}}+...+\frac{1}{(n-1)^2}+\frac{1}{n^2}< 1\) ;

b, \(\text{B}=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\).

 

Answer 1

\(\frac{1}{2^2}< \frac{1}{1.2}\)

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

........

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

=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

Đpcm 

Answer 2

b)B=1/4(1/2^2+1/3^2+...+1/n^2)=1/4*A<1/4

Answer 3

tìm 2 số 3:

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