Question

Cho \(A=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+98}\)

Giải thích tại sao \(A< \dfrac{2}{3}\)

Answer 1

Áp dụng công thức: \(1+2+3+...+n=\dfrac{n+\left(n+1\right)}{2}\) ta có:

\(A=\dfrac{2}{2.3}+\dfrac{2}{4.5}+...+\dfrac{2}{98.99}=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=\dfrac{64}{99}< \dfrac{66}{99}=\dfrac{2}{3}\)