Question

[abcdefghijklmnopqrstuvwxyz]

Tính biểu thức sau: 

\(\frac{1+2}{2}\)+ \(\frac{1+2+3}{3}\)+ \(\frac{1+2+3+4}{4}\)+ ....... + \(\frac{1+2+3+....+197+198+199}{199}\)

Answer 1

[Phạm Tuấn Đạt]

\(\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)

\(=\frac{\frac{3.2}{2}}{2}+\frac{\frac{4.3}{2}}{3}+...+\frac{\frac{200.199}{2}}{199}\)

\(=\frac{3.2}{2}.\frac{1}{2}+\frac{4.3}{2}.\frac{1}{3}+...+\frac{200.199}{2}.\frac{1}{199}\)

\(=\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)

\(=\frac{3+4+...+200}{2}=\frac{203.198}{2}.\frac{1}{2}=\frac{20097}{2}\)