题目

Problem

(a) Find, in terms of $a$ , the first 3 terms, in ascending powers of $x$ , of the binomial expansion of

(2+ax)^6

where $a$ is a non-zero constant. Give each term in simplest form.

(3)

f(x)=\left(3+\frac1x\right)^2(2+ax)^6

Given that the constant term in the expansion of $f(x)$ is $576$

(b) find the value of $a$ .

(4)

解答

展开

只需要前三项，即 $x^0,x^1,x^2$ 的项。二项式系数分别是 $1,6,15$ 。

展开

\begin{align*} (2+ax)^6 =&\,2^6+{6\choose1}2^5(ax) +{6\choose2}2^4(ax)^2+\cdots\\ =&\,64+192ax+240a^2x^2+\cdots . \end{align*}

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要找常数项，需要把 $\left(3+\frac1x\right)^2$ 展开后，与 (a) 中不同幂次的项配对，使 $x$ 的总次数为 $0$ 。

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First,

\begin{align*} \left(3+\frac1x\right)^2 =&\,9+\frac6x+\frac1{x^2}. \end{align*}

Also,

\begin{align*} (2+ax)^6=64+192ax+240a^2x^2+\cdots . \end{align*}

The constant term in $f(x)$ is formed by

9\cdot64,\qquad \frac6x\cdot192ax,\qquad \frac1{x^2}\cdot240a^2x^2.

So the constant term is

\begin{align*} 9(64)+6(192a)+240a^2 =&\,576+1152a+240a^2. \end{align*}

This is equal to $576$ , so

\begin{align*} 576+1152a+240a^2=&\,576\\ 1152a+240a^2=&\,0\\ a(1152+240a)=&\,0. \end{align*}

Since $a$ is non-zero,

\begin{align*} 1152+240a=&\,0\\ a=&\,-\frac{1152}{240}\\ =&\,-\frac{24}{5}. \end{align*}