Đáp án:
\[n = 12\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{a_0} = C_n^0{.2^n}\\
{a_1} = C_n^1{.2^{n - 1}}{.3^1}\\
{a_2} = C_n^2{.2^{n - 2}}{.3^2}\\
.....\\
{a_n} = C_n^n{.3^n}
\end{array} \right.\\
{3^n}{a_0} + {3^{n - 1}}.{a_1} + ..... + {a_n} = {3^{n + 12}}\\
\Leftrightarrow C_n^0{.2^n}{.3^n} + C_n^1{.2^{n - 1}}{.3^n} + C_n^2{.2^{n - 2}}{.3^n} + ..... + C_n^n{.2^0}{.3^n} = {3^{n + 12}}\\
\Leftrightarrow {3^n}.\left( {C_n^0{{.2}^n} + C_n^1{{.2}^{n - 1}} + C_n^2{{.2}^{n - 2}} + ..... + C_n^n{{.2}^0}} \right) = {3^{n + 12}}\\
\Leftrightarrow {3^n}.{\left( {2 + 1} \right)^n} = {3^{n + 12}}\\
\Leftrightarrow {3^{2n}} = {3^{n + 12}}\\
\Leftrightarrow 2n = n + 12\\
\Leftrightarrow n = 12
\end{array}\)
