Question

1/(1+2)+1/(1+2+3)+.....+1/1+2+3+.....+2014 = ?

Answer 1

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

\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2014.2015}=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)

    \(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)=2.\left(\frac{1}{2}-\frac{1}{2015}\right)=1-\frac{2}{2015}=\frac{2013}{2015}\)