2020-M2-01 – Solving Master

答案：(a)

1 - 4 x + 6 x^{2} - 4 x^{3} + x^{4}

(b)

\frac{1}{2}

利用二項式定理，可得
$\begin{array}{cl} (1 - x)^{4} \\ = & C_{0}^{4} + C_{1}^{4} (- x) + C_{2}^{4} (- x)^{2} + C_{3}^{4} (- x)^{3} + C_{4}^{4} (- x)^{4} \\ = & 1 - 4 x + 6 x^{2} - 4 x^{3} + x^{4} \end{array}$
$\begin{array}{cl} (1 + k x)^{9} (1 - x)^{4} \\ = & (C_{0}^{9} + C_{1}^{9} (k x) + C_{2}^{9} (k x)^{2} + \dots) (1 - 4 x + 6 x^{2} - 4 x^{3} + x^{4}) \end{array}$

考慮 $x^{2}$ 的係數，可得

$\begin{array}{rcl} C_{0}^{9} (6) + C_{1}^{9} (k) (- 4) + C_{2}^{9} (k)^{2} (1) & = & - 3 \\ 6 - 36 k + 36 k^{2} & = & - 3 \\ 4 k^{2} - 4 k + 1 & = & 0 \\ (2 k - 1)^{2} & = & 0 \end{array}$

$∴ k = \frac{1}{2}$ (重根)。