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a) Ta có: $ {(1 + 2x + 3{x^2})^{10}} = {\left[ {\left( {1 + 2x} \right) + 3{x^2}} \right]^{10}} $
$ = \,C_{10}^0{\left( {1 + 2x} \right)^{10}} + C_{10}^1{\left( {1 + 2x} \right)^9}3{x^2} + C_{10}^2{\left( {1 + 2x} \right)^8}{\left( {3{x^2}} \right)^2} + ... + C_{10}^{10}{\left( {3{x^2}} \right)^{10}} $ $ \begin{array}{l}
= \,\,\,\,C_{10}^0\left[ {C_{10}^0 + C_{10}^12x + C_{10}^2{{\left( {2x} \right)}^2} + C_{10}^3{{\left( {2x} \right)}^3} + ... + C_{10}^{10}{{\left( {2x} \right)}^{10}}} \right] + \\
+ \,\,\,C_{10}^0\left[ {C_9^12x + C_9^2{{\left( {2x} \right)}^2} + ... + C_9^9{{\left( {2x} \right)}^9}} \right]3{x^2}\\
+ \,\,C_{10}^0\left[ {C_8^0 + ... + C_8^8{{\left( {2x} \right)}^8}} \right]9{x^4} + ... + C_{10}^{10}{\left( {3{x^2}} \right)^{10}}
\end{array} $
Do hệ số của số hạng chứa $ {x^4} $ là:
$ \begin{array}{l}
{a_4} = \,\,\,{2^4}C_{10}^0C_{10}^4 + {3.2^2}C_{10}^1C_9^2 + 9C_{10}^2C_8^0
= \,\,\,16.1.210 + 12.10.36 + 9.45.1,\\
\Rightarrow \,\,\,\,\,{a_4}\, = \,\,\,\,3360 + 4320 + 405 = 8085
\end{array} $
b) Khi $ x = 1,\, $ ta có: $ {\left( {1 + 2 + 3} \right)^{10}} = {a_0} + {a_1}+{a_2} + ... + {a_{20}} =6^{10}$
Vậy
$ S = 6^{10}$
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