tính tổng: S= 2020 C 0 +4(2020 C 1) +7(2020 C 2) +….+ (3. 2020+1) 2020 C 2020

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1. 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:

   \(\begin{array}{l}
   S = C_0^{2020} + 4.C_{2020}^1 + 7.C_{2020}^2 + …. + \left( {3.2020 + 1} \right)C_{2020}^{2020}\\
    = 3.\left( {1.C_{2020}^1 + 2.C_{2020}^2 + 3.C_{2020}^3 + …. + 2020.C_{2020}^{2020}} \right) + \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + …. + C_{2020}^{2020}} \right)\\
    = 3.\left( {2020.C_{2019}^0 + 2020C_{2019}^1 + 2020.C_{2019}^2 + …. + 2020.C_{2019}^{2019}} \right) + \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + …. + C_{2020}^{2020}} \right)\\
    = 6060\left( {C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + …. + C_{2019}^{2019}} \right) + \left( {C_{2020}^0 + C_{2020}^1 + C_{2020}^2 + …. + C_{2020}^{2020}} \right)\\
    = 6060.{\left( {1 + 1} \right)^{2019}} + {\left( {1 + 1} \right)^{2020}}\\
    = {6060.2^{2019}} + {2^{2020}}\\
    = {6062.2^{2019}}
   \end{array}\)

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