Question

Cho B = \(\dfrac{1}{4}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{19}\). Hãy chứng minh rằng B > 1

Mn help mk vs ạk (Mai thi oy)

Answer 1

\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)\)

Vì \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\) nên \(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{5}{9}>\dfrac{1}{2}\).

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+...+\dfrac{1}{19}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{10}{19}>\dfrac{1}{2}\).

\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}>1\)

\(\Rightarrow B>1\)

Answer 2

B=\(\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\right)>\dfrac{1}{4}+15nhân\dfrac{1}{20}\)

B>\(>\dfrac{1}{4}+\dfrac{15}{20}=\dfrac{1}{4}+\dfrac{3}{4}=1\)

Suy ra B>1