Ta có:
\(1+2+3=\frac{\left(1+3\right).3}{2}=\frac{4.3}{2}\)
\(1+2+3+4=\frac{\left(1+4\right).4}{2}=\frac{5.4}{2}\)
\(.................\)
\(1+2+3+...+59=\frac{\left(1+59\right).59}{2}=\frac{60.59}{2}\)
Nên : \(M=\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+...+\frac{1}{\frac{60.59}{2}}\) , suy ra :\(2M=\frac{1}{4.3}+\frac{1}{5.4}+...+\frac{1}{60.59}\)
\(2M=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\)
\(2M=\frac{1}{3}-\frac{1}{60}<\frac{1}{3}\)
Do đó : \(M<\frac{1}{3}.2\)
\(M<\frac{2}{3}\)
Mình sửa lại \(2M\) thành 1/2 M nhé
Xét Tổng \(A=1+2+3+4+......+n\)Ta có: \(A=\frac{n\left(n+1\right)}{2}\)Vậy \(\frac{1}{A}=\frac{2}{n\left(n+1\right)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Vậy \(\frac{1}{1+2+3}=\frac{1}{6}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
\(\frac{1}{1+2+3+4}=\frac{1}{10}=2\left(\frac{1}{4}-\frac{1}{5}\right)\)
Tương tự:....................................
\(\frac{1}{1+2+3+4+.......+59}=\frac{1}{1770}=2\left(\frac{1}{59}-\frac{1}{60}\right)\)
Vậy \(M=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.................+\frac{1}{59}-\frac{1}{60}\right)\)Hay \(M=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}<\frac{2}{3}\)