Question

Cho \(S=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...........+\dfrac{1}{20}\) Hãy so sánh \(S\) và \(\dfrac{7}{12}\)

Help me!!!!!!!!!

Answer 1

Theo đề bài :

\(S=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)

S có tất cả 10 hạng tử, do đó :

\(S\) > \(\left(\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}\right)+\left(\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}\right)\)

\(S\) > \(5\times\dfrac{1}{15}+5\times\dfrac{1}{20}=\dfrac{7}{12}\)

Vậy \(S>\dfrac{7}{12}\)