Đáp án:
D
Giải thích các bước giải:
\(\begin{array}{l}S = 3C_n^0 + 3C_n^1 + 3C_n^2 + ... + 3C_n^n + C_n^1 + 2C_n^2 + ... + nC_n^n\\\,\,\,\, = 3\left( {C_n^0 + C_n^1 + C_n^2 + ... + C_n^n} \right) + \left( {C_n^1 + 2C_n^2 + ... + nC_n^n} \right)\\\,\,\,\,\, = 3{S_1} + {S_2}\end{array}\)
\(\begin{array}{l}{S_1} = C_n^0 + C_n^1 + C_n^2 + ... + C_n^n = {2^n}\\{S_2} = C_n^1 + 2C_n^2 + ... + nC_n^n\end{array}\)
Xét SHTQ
\(\begin{array}{l}kC_n^k = k\frac{{n!}}{{k!\left( {n - k} \right)!}} = \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k} \right)!}}\\\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)!}}{{\left( {k - 1} \right)!\left( {n - k} \right)!}} = nC_{n - 1}^{k - 1}\end{array}\)
\(\begin{array}{l} \Rightarrow {S_2} = C_n^1 + 2C_n^2 + ... + nC_n^n\\ \Rightarrow {S_2} = nC_{n - 1}^0 + nC_{n - 1}^1 + nC_{n - 1}^2 + ... + nC_{n - 1}^{n - 1}\\\,\,\,\,\,\,{S_2} = n\left( {C_{n - 1}^0 + C_{n - 1}^1 + C_{n - 1}^2 + ... + C_{n - 1}^{n - 1}} \right)\\\,\,\,\,\,\,{S_2} = n{2^{n - 1}}\end{array}\)
\(\begin{array}{l} \Rightarrow S = {3.2^n} + n{2^{n - 1}} = 3840\\ \Leftrightarrow n = 9\end{array}\)
\( \Rightarrow {\left( {1 + x - {x^2} + {x^3}} \right)^n} = {\left( {1 + x - {x^2} + {x^3}} \right)^9}\)
Thay \(x = 1\) ta có tổng tất cả các hệ số của khai triển trên là \({\left( {1 + 1 - 1 + 1} \right)^9} = {2^9}\).
Chọn D.
