Đáp án:
\(S = \dfrac{1}{{2020}}\left( {{2^{2020}} - 1} \right)\).
Giải thích các bước giải:
Ta có:
\(\dfrac{1}{{k + 1}}C_n^k = \dfrac{1}{{k + 1}}\dfrac{{n!}}{{k!\left( {n - k!} \right)}} = \dfrac{1}{{n + 1}}\dfrac{{\left( {n + 1} \right)!}}{{\left( {k + 1} \right)!\left( {n - k} \right)!}} = \dfrac{1}{{n + 1}}C_{n + 1}^{k + 1}\)
Do đó:
\(\begin{array}{l}C_{2019}^0 = \dfrac{1}{{2020}}C_{2020}^1\\\dfrac{1}{2}C_{2019}^1 = \dfrac{1}{{2020}}C_{2020}^2\\\dfrac{1}{3}C_{2019}^2 = \dfrac{1}{{2020}}C_{2020}^3\\...\\\dfrac{1}{{2020}}C_{2019}^{2019} = \dfrac{1}{{2020}}C_{2020}^{2020}\end{array}\)
\(\begin{array}{l} \Rightarrow S = \dfrac{1}{{2020}}C_{2020}^1 + \dfrac{1}{{2020}}C_{2020}^2 + \dfrac{1}{{2020}}C_{2020}^3 + ... + \dfrac{1}{{2020}}C_{2020}^{2020}\\\,\,\,\,\,\,S = \dfrac{1}{{2020}}\left( {C_{2020}^1 + C_{2020}^2 + C_{2020}^3 + ... + C_{2020}^{2020}} \right)\\\,\,\,\,\,\,S = \dfrac{1}{{2020}}\left[ {{{\left( {1 + 1} \right)}^{2020}} - C_{2020}^0} \right]\\\,\,\,\,\,\,S = \dfrac{1}{{2020}}\left( {{2^{2020}} - 1} \right)\end{array}\)
Vậy \(S = \dfrac{1}{{2020}}\left( {{2^{2020}} - 1} \right)\).
