Question

Tính tổng S

S = \(\frac{1}{1+2}+\frac{1}{1+2+3}+......+\frac{1}{1+2+3+.....+2017}\)

Answer 1

\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2017}\)

\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2017.2018}\)

\(\frac{1}{2}S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{2018}\)

\(\frac{1}{2}S=\frac{504}{1009}\)

=> \(S=\frac{1008}{1009}\)