Question

[Trần Nguyễn Duy Long]

Bài 1 Cho A = 4 / 15.19+ 4/ 19.23 +.......+ 4/ 399.403 .

Chứng minh 16/81 < A < 16/80

Answer 1

[Nguyễn Lê Phước Thịnh]

\(A=\dfrac{4}{15\cdot19}+\dfrac{4}{19\cdot23}+...+\dfrac{4}{399\cdot403}\)

\(=\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+...+\dfrac{1}{399}-\dfrac{1}{403}\)

\(=\dfrac{1}{15}-\dfrac{1}{403}=\dfrac{403-15}{403\cdot15}=\dfrac{388}{6045}\)

\(\dfrac{16}{81}=\dfrac{16\cdot24,25}{81\cdot24,25}=\dfrac{388}{1964,25}\)

\(\dfrac{16}{80}=\dfrac{16\cdot24,25}{80\cdot24,25}=\dfrac{388}{1940}\)

Vì 1940<1964,25<6045

nên \(\dfrac{16}{80}>\dfrac{16}{81}>A\)