Question

1/1+2 +1/1+2+3+...+1/1+2+3+...+2016

Answer 1

Ta có: \(\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+2016}\)

\(=\frac{1}{2\times\frac32}+\frac{1}{3\times\frac42}+\cdots+\frac{1}{2016\times\frac{2017}{2}}\)

\(=\frac{2}{2\times3}+\frac{2}{3\times4}+\cdots+\frac{2}{2016\times2017}\)

\(=2\times\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2\times\left(\frac12-\frac{1}{2017}\right)=1-\frac{2}{2017}=\frac{2015}{2017}\)