Question

CMR: \(\dfrac{1}{3}.\dfrac{4}{6}.\dfrac{7}{9}.\dfrac{10}{12}...\dfrac{208}{210}< \dfrac{1}{25}\)

Answer 1

Có:

\(\dfrac{n}{n+2}< \dfrac{n-1}{n}\)(Vì
\(n^2< n^2+n-2\forall n>2\))

Nên ta có

\(F=\dfrac{1}{3}.\dfrac{4}{6}....\dfrac{208}{201}\)

\(\Rightarrow F< \dfrac{1}{3}.\dfrac{3}{4}.\dfrac{6}{7}...\dfrac{207}{208}\)

\(\Rightarrow F^2< \dfrac{1.4.7...208}{3.6.9.12...210}.\dfrac{1.3.6.9...207}{3.4.7.10.208}\)

\(\Rightarrow F^2=\dfrac{1}{210}.\dfrac{1}{3}\)

\(\Rightarrow F^2=\dfrac{1}{630}< \left(\dfrac{1}{25}\right)^2\)

Vậy F\(< \dfrac{1}{25}\)