Question

Chứng minh rằng

4^2/20*24+4^2/24*28+...+4^2/76*80<1

Answer 1

Đặt A=\(\dfrac{4^2}{20.24}+\dfrac{4^2}{24.28}+...+\dfrac{4^2}{76.80}\)

A=\(\dfrac{16}{20.24}+\dfrac{16}{24.28}+...+\dfrac{16}{76.80}\)

A=4.[\(\dfrac{4}{20.24}+\dfrac{4}{24.28}+...+\dfrac{4}{76.80}\)]

A=4.\(\left[\dfrac{1}{20}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{28}+...+\dfrac{1}{76}-\dfrac{1}{80}\right]\)

A=4.\(\left[\dfrac{1}{20}+\dfrac{1}{24}-\dfrac{1}{24}+\dfrac{1}{28}-\dfrac{1}{28}+...+\dfrac{1}{78}-\dfrac{1}{78}-\dfrac{1}{80}\right]\)

A=4.\(\left[\dfrac{1}{20}-\dfrac{1}{80}\right]\)

A=4.\(\dfrac{3}{80}\)

A=\(\dfrac{3}{20}\)<1

=>A<1

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