Giải thích các bước giải:
Câu 3:
Ta có:
$\begin{array}{l}
3C_n^3 - 2A_n^2 = n\left( {n - 4} \right)\left( {Do:n \ge 3} \right)\\
\Leftrightarrow 3.\frac{{n!}}{{\left( {n - 3} \right)!.3!}} - 2.\frac{{n!}}{{\left( {n - 2} \right)!}} = n\left( {n - 4} \right)\\
\Leftrightarrow \frac{1}{2}\left( {n - 2} \right)\left( {n - 1} \right)n - 2\left( {n - 1} \right)n - n\left( {n - 4} \right) = 0\\
\Leftrightarrow n\left[ {\frac{1}{2}\left( {n - 2} \right)\left( {n - 1} \right) - 2\left( {n - 1} \right) - \left( {n - 4} \right)} \right] = 0\\
\Leftrightarrow \frac{1}{2}\left( {n - 2} \right)\left( {n - 1} \right) - 2\left( {n - 1} \right) - \left( {n - 4} \right) = 0\left( {Do:n \ge 3} \right)\\
\Leftrightarrow \frac{1}{2}\left( {{n^2} - 3n + 2} \right) - 3n + 6 = 0\\
\Leftrightarrow {n^2} - 9n + 14 = 0\\
\Leftrightarrow \left( {n - 2} \right)\left( {n - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
n = 2\left( l \right)\\
n = 7\left( c \right)
\end{array} \right.\\
\Leftrightarrow n = 7
\end{array}$
Vậy $n = 7$
Câu 4:
$\begin{array}{l}
S = 9C_{20}^1 + {9^2}C_{20}^2 + ... + {9^{19}}C_{20}^{19} + {9^{20}}C_{20}^{20}\\
\Leftrightarrow S + C_{20}^0 = C_{20}^0 + C_{20}^19 + C_{20}^2{9^2} + ... + C_{20}^{19}{9^{19}} + C_{20}^{20}{9^{20}}\\
\Leftrightarrow S + 1 = C_{20}^0{.9^0}{.1^{20}} + C_{20}^1{.9^1}{.1^{19}} + C_{20}^2{.9^2}{.1^{18}} + ... + C_{20}^{19}{.9^{19}}{.1^1} + C_{20}^{20}{.9^{20}}{.1^0}\\
\Leftrightarrow S + 1 = {\left( {9 + 1} \right)^{20}}\\
\Leftrightarrow S + 1 = {10^{20}}\\
\Leftrightarrow S = {10^{20}} - 1
\end{array}$
Vậy $S = {10^{20}} - 1$
