Đáp án:
\({6062.2^{2019}}\)
Giải thích các bước giải:
Ta có công thức tổng quát
\(k.C_n^k = k.\dfrac{{n!}}{{k!\left( {n - k} \right)!}} = \dfrac{{n!}}{{\left( {k - 1} \right)!.\left( {n - k} \right)!}} \\= n.\dfrac{{\left( {n - 1} \right)!}}{{\left( {k - 1} \right)!.\left[ {\left( {n - 1} \right) - \left( {k - 1} \right)} \right]!}} \\= n.C_{n - 1}^{k - 1}\)
⇒
\(\begin{array}{l} S = C_0^{2020} + 4.C_{2020}^1 + 7.C_{2020}^2 + .... + \left( {3.2020 + 1} \right)C_{2020}^{2020}\\ = 3.\left( {1.C_{2020}^1 + 2.C_{2020}^2 + 3.C_{2020}^3 + .... + 2020.C_{2020}^{2020}} \right) \\+ \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + .... + C_{2020}^{2020}} \right)\\ = 3.\left( {2020.C_{2019}^0 + 2020C_{2019}^1 + 2020.C_{2019}^2 + .... + 2020.C_{2019}^{2019}} \right) \\+ \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + .... + C_{2020}^{2020}} \right)\\ = 6060\left( {C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + .... + C_{2019}^{2019}} \right)\\ + \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + .... + C_{2020}^{2020}} \right)\\ = 6060.{\left( {1 + 1} \right)^{2019}} + {\left( {1 + 1} \right)^{2020}}\\ = {6060.2^{2019}} + {2^{2020}}\\ = {6062.2^{2019}} \end{array}\)
