题目

Problem

(a) Find the first 4 terms, in ascending powers of $x$ , of the binomial expansion of

\begin{align*} (3+2x)^6 \end{align*}

giving each coefficient in simplest form.

(4)

(b) Hence find the coefficient of $x^2$ in the expansion of

\begin{align*} \left(2x^2-\frac{1}{6x}\right)(3+2x)^6 \end{align*}

(3)

解答

展开

用二项式定理 $(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ ，逐项计算 $k=0,1,2,3$ 。

展开

Using the binomial theorem with $a=3$ , $b=2x$ and $n=6$ ,

\begin{align*} (3+2x)^6=\sum_{k=0}^{6}\binom{6}{k}3^{6-k}(2x)^k. \end{align*}

The first four terms ( $k=0,1,2,3$ ) are:

\begin{align*} k=0:\quad&\binom{6}{0}3^6=1\times729=729\\[2mm] k=1:\quad&\binom{6}{1}3^5(2x)=6\times243\times2x=2916x\\[2mm] k=2:\quad&\binom{6}{2}3^4(2x)^2=15\times81\times4x^2=4860x^2\\[2mm] k=3:\quad&\binom{6}{3}3^3(2x)^3=20\times27\times8x^3=4320x^3 \end{align*}

Therefore

\begin{align*} (3+2x)^6=729+2916x+4860x^2+4320x^3+\cdots \end{align*}

展开

把 $(2x^2-\frac{1}{6x})$ 分别乘以 (a) 的展开式，找出所有产生 $x^2$ 的项。

展开

From part (a),

\begin{align*} (3+2x)^6=729+2916x+4860x^2+4320x^3+\cdots \end{align*}

Now

\begin{align*} \left(2x^2-\frac{1}{6x}\right)(3+2x)^6=2x^2(3+2x)^6-\frac{1}{6x}(3+2x)^6. \end{align*}

For the coefficient of $x^2$ :

From $2x^2\times(3+2x)^6$ : the term $2x^2\times729$ gives $x^2$ .
From $-\dfrac{1}{6x}\times(3+2x)^6$ : the term $-\dfrac{1}{6x}\times4860x^2$ gives $x^2$ , and $-\dfrac{1}{6x}\times2916x$ gives $x^0$ (not $x^2$ ).

Wait, let me be more careful. We need the total power of $x$ to be 2:

Therefore the coefficient of $x^2$ is

\begin{align*} 2\times729+\left(-\frac{1}{6}\right)\times4320=1458-720=738. \end{align*}