Question

Thực hiện phép tính sau :

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

Ai nhanh và đúng tick nè

Answer 1

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

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

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

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

\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)

\(=76\)

Answer 2

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

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

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

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

          \(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)

            \(=\frac{15+135+2}{2}\)

              \(=\frac{152}{2}\)\(=76\)