Question

Chung minh A<\(\dfrac{1}{10}\) biet :

A=\(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+......+\dfrac{1}{252.509}\)

Answer 1

\(\dfrac{5}{2}A=\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{504.509}\)

\(\dfrac{5}{2}A=\dfrac{9-4}{4.9}+\dfrac{14-9}{9.14}+\dfrac{19-14}{14.19}+...+\dfrac{509-504}{504.509}\)

\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\)

\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{509}\)

\(A=\left(\dfrac{1}{4}-\dfrac{1}{509}\right).\dfrac{2}{5}\)

\(A=\dfrac{1}{10}-\dfrac{2}{2545}< \dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{1}{10}\)(đpcm)

Chúc bạn học tốt!

Answer 2

Ta có:

A=\(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)

A=2.(\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{504.509}\))

A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\))

A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{509}\))

A=\(\dfrac{2}{5}\).(\(\dfrac{509}{2036}-\dfrac{4}{2036}\))

A=\(\dfrac{2}{5}\).\(\dfrac{505}{2036}\)

A=\(\dfrac{101}{1018}\)