A=1+\(\frac{1+2}{2}\)+\(\frac{1+2+3}{3}\)+...+\(\frac{1+2+3+...+16}{16}\)

A=\(\frac{2}{2}\)+\(\frac{3}{2}\)+\(\frac{4}{2}\)+...+\(\frac{17}{2}\)

A=\(\frac{2+3+4+...+17}{2}\)

A=76(đề thi HSG huyện tui có tui làm zậy mà cũng có điểm tuyệt đối)

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+3+....+16\right)\)

\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+\frac{1}{4}\cdot\frac{4.5}{2}+.....+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}\)

\(A=\frac{\left(2+3+4+.....+17\right)}{2}=\frac{\left(2+17\right).16}{2}=\frac{152}{2}=76\)

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)

\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{16}\left(1+2+3+....+16\right)\)

\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+....+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{17}{2}\)

\(A=\frac{\left(2+3+4+....+17\right)}{2}=\frac{\left(2+17\right).\left(17-2+1\right):2}{2}=\frac{152}{2}=76\)

Ta có: \(A = 1+{1+2\over 2} + {1+2+3\over 3} +...+{1+2+...+ 16\over 16}\)

Xét: \(S_n = 1+2+3+...+n =\frac{n(n+1)}{n} (n \in N^*)\)

=> \({S_n\over n} = {(n+1)\over 2}\)

Thay vào biểu thức A ta có: 

\(A=1 + {3\over 2} + {4\over 2} + ... + {17\over 2}\)

\(A={(2+3+4+...+17)\over 2}\)

\(A={(17+2)[(17-2+1):2]\over 2} = {152\over2}=76\)

\(A=1+\frac{1+2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)

\(A=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+16}{2}=1+\frac{\left(1+1+...+1\right)+\left(2+3+...+16\right)}{2}\)

\(A=1+\frac{15+\frac{\left(2+16\right).15}{2}}{2}=1+\frac{150}{2}=76\)