题目

Problem

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

(1-4x)^7

giving each term in simplest form.

(4)

In the series expansion of

(5+kx)(1-4x)^7

where $k$ is a constant

the coefficient of the term in $x^2$ is $1316$

(b) Use the answer to part (a) to find the value of $k$ .

(2)

解答

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使用 binomial expansion。题目要前 $4$ 项，所以取 $r=0,1,2,3$ 的项。

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Using the binomial expansion,

\begin{align*} (1-4x)^7 =&\,1+7(-4x)+\binom72(-4x)^2\\ &\,\hspace{2pt}+\binom73(-4x)^3+\cdots \end{align*}

Now simplify each term:

\begin{align*} (1-4x)^7 =&\,1-28x+21(16x^2)\\ &\,\hspace{2pt}+35(-64x^3)+\cdots\\ =&\,1-28x+336x^2-2240x^3+\cdots. \end{align*}

Therefore, the first $4$ terms are

\begin{align*} 1-28x+336x^2-2240x^3. \end{align*}

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要找 $(5+kx)(1-4x)^7$ 中 $x^2$ 的系数。它来自两部分： $5$ 乘上原展开里的 $x^2$ 项，以及 $kx$ 乘上原展开里的 $x$ 项。

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From part (a),

\begin{align*} (1-4x)^7=1-28x+336x^2+\cdots. \end{align*}

\begin{align*} (5+kx)(1-4x)^7, \end{align*}

the coefficient of $x^2$ is

\begin{align*} 5(336)+k(-28). \end{align*}

\begin{align*} 5(336)-28k=&\,1316\\ 1680-28k=&\,1316\\ 28k=&\,364\\ k=&\,13. \end{align*}