Question

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

Answer 1

\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2017}.\left(1+2+3+...+2017\right)\)

\(S=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{2017}.\frac{\left(1+2017\right).2017}{2}\)

\(S=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2018}{2}\)

\(S=\frac{1}{2}.\left(2+3+4+...+2018\right)\)

\(S=\frac{1}{2}.\frac{\left(2+2018\right).2017}{2}\)

\(S=\frac{2020.2017}{4}=505.2017=1018585\)