题目

Problem

The binomial expansion, in ascending powers of $x$ , of $(3 + px)^5$

where $p$ is a constant, can be written in the form

$A + Bx + Cx^2 + Dx^3 + \cdots$

where $A$ , $B$ , $C$ and $D$ are constants.

(a) Find the value of $A$

Given that $B = 18D$ and $p < 0$ ,

(b) find (i) the value of $p$ (ii) the value of $C$

(7)

题目中文翻译

$(3+px)^5$ 的升幂二项展开式可写成 $A+Bx+Cx^2+Dx^3+\cdots$ ，其中 $A,B,C,D$ 为常数。

(a) 求 $A$ 的值

已知 $B=18D$ 且 $p<0$ ，

(b) (i) 求 $p$ 的值 (ii) 求 $C$ 的值

（7分）

解答

二项展开 $(a+b)^n$ 的常数项为 $a^n$ 。写出 $B$ 和 $D$ 关于 $p$ 的表达式，利用 $B=18D$ 列方程。注意 $p<0$ 。

(a) The constant term is:

$A = 3^5 = \boxed{243}$

(b)(i) The binomial expansion of $(3 + px)^5$ :

$(3 + px)^5 = \sum_{r=0}^{5} \binom{5}{r} 3^{5-r}(px)^r$

Coefficient of $x$ :

$B = \binom{5}{1} \cdot 3^4 \cdot p = 5 \times 81 \times p = 405p$

Coefficient of $x^3$ :

$D = \binom{5}{3} \cdot 3^2 \cdot p^3 = 10 \times 9 \times p^3 = 90p^3$

Given $B = 18D$ :

405p =&\, 18 \times 90p^3 \\[4mm] 405p =&\, 1620p^3 \\[4mm] 405p - 1620p^3 =&\, 0 \\[4mm] 405p(1 - 4p^2) =&\, 0 \end{align*}$$ Since $p \neq 0$: $$1 - 4p^2 = 0 \implies p^2 = \frac{1}{4} \implies p = \pm\frac{1}{2}$$ Given $p < 0$: $$\boxed{p = -\frac{1}{2}}$$ **(b)(ii)** Coefficient of $x^2$: $$C = \binom{5}{2} \cdot 3^3 \cdot p^2 = 10 \times 27 \times p^2 = 270p^2$$ $$C = 270 \times \left(-\frac{1}{2}\right)^2 = 270 \times \frac{1}{4} = \boxed{67.5}$$