Question

[Lê Hiển Vinh]

Cho: 

\(\frac{1}{m}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+59}\)

Chứng minh rằng: \(m>\frac{2}{3}\).

Answer 1

[Hoàng Lê Bảo Ngọc]

Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)

\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)

Answer 2

[Ngô Tấn Đạt]

\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)

\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)