题目

Problem

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

\left(1-\frac16x\right)^9

giving each term in simplest form.

(3)

(b) Hence find the coefficient of $x^3$ in the expansion of

(10x+3)\left(1-\frac16x\right)^9

giving the answer in simplest form.

(2)

解答

展开

用二项式展开的前四项。这里的“小量”是 $-\frac16x$ ，所以符号会正负交替。

展开

Using the binomial expansion,

\begin{align*} \left(1-\frac16x\right)^9 =&\,1+9\left(-\frac16x\right) +\binom92\left(-\frac16x\right)^2\\ &\,\hspace{2pt} +\binom93\left(-\frac16x\right)^3+\cdots\\ =&\,1-\frac32x+x^2-\frac7{18}x^3+\cdots. \end{align*}

So the first four terms are

\begin{align*} 1-\frac32x+x^2-\frac7{18}x^3. \end{align*}

展开

要得到 $x^3$ ，有两种来源： $10x$ 乘上括号里的 $x^2$ 项，或 $3$ 乘上括号里的 $x^3$ 项。

展开

From part (a),

\left(1-\frac16x\right)^9 =1-\frac32x+x^2-\frac7{18}x^3+\cdots.

The coefficient of $x^3$ in

\begin{align*} (10x+3)\left(1-\frac16x\right)^9 \end{align*}

\begin{align*} 10(1)+3\left(-\frac7{18}\right) =&\,10-\frac76\\ =&\,\frac{60}{6}-\frac76\\ =&\,\frac{53}{6}. \end{align*}