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