Question

Cho \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

Chứng minh A < \(\frac{1}{2}\)

Answer 1

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)

=> \(A< 2\left(đpcm\right)\)

Answer 2

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)

\(A< 2\left(đpcm\right)\)

Answer 3

\(\Rightarrow\)A<2(dpcm)

Answer 4

đpcm là gì các bạn

Answer 5

deo biet

Answer 6

dieu phaui chung minh

Answer 7

A<2(dpcm)

Answer 8

Ko bít 

Answer 9

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