Question

Tính:

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+......+\frac{1}{20}\left(1+2+3+....+20\right)\)

Answer 1

Ta đã biết: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)

Ta có: \(A=1+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+...+\frac{1}{20}.\left(\frac{20.21}{2}\right)\)

\(A=1+\frac{3}{2}+\frac{4}{2}+....+\frac{21}{2}\)

\(A=\frac{1}{2}.\left(2+3+....+21\right)\)

Tổng trong ngoặc có:21-2+2=20 (số hạng)

\(=>A=\frac{1}{2}.\left(\frac{\left(21+2\right).20}{2}\right)=\frac{1}{2}.230=115\)

Vậy..........

Answer 2

Nể Hoàng Phúc giải nhanh thế !!!!