\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{17\cdot21}< 1\)
\(A=\dfrac{4}{4}\cdot\left(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{17\cdot21}\right)< 1\)
\(A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}< 1\)
\(A=1-\dfrac{1}{21}< 1\) (đúng) (đpcm).
Ta có: \(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)
=\(\dfrac{4}{4}.\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{17.21}\right)\)
=\(1\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}\right)\)
=\(1-\dfrac{1}{21}\)
mà \(1-\dfrac{1}{21}\)<1
=>\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)<1
