题目

Problem

The binomial expansion, in ascending powers of $x$ , of $(1+kx)^n$ is

1-24x+270x^2+px^3+\cdots

where $k$ and $p$ are constants and $n$ is an integer greater than $1$

(a) Use this information to set up and solve two simultaneous equations to find the value of $k$ and the value of $n$ .

(6)

(b) Hence find the value of $p$ .

(2)

解答

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比较 $(1+kx)^n$ 的 $x$ 项和 $x^2$ 项系数。第一项给 $nk=-24$ ，第二项给 $\binom n2k^2=270$ 。

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For

\begin{align*} (1+kx)^n, \end{align*}

the coefficient of $x$ is

nk.

\begin{align*} nk=-24. \end{align*}

The coefficient of $x^2$ is

\begin{align*} \binom n2k^2=\frac{n(n-1)}2k^2. \end{align*}

\begin{align*} \frac{n(n-1)}2k^2=270. \end{align*}

From $nk=-24$ ,

\begin{align*} k=-\frac{24}{n}. \end{align*}

Substitute into the second equation:

\begin{align*} \frac{n(n-1)}2\left(-\frac{24}{n}\right)^2=&\,270\\ \frac{n(n-1)}2\cdot\frac{576}{n^2}=&\,270\\ \frac{288(n-1)}{n}=&\,270\\ 288n-288=&\,270n\\ 18n=&\,288\\ n=&\,16. \end{align*}

Then

\begin{align*} 16k=&\,-24\\ k=&\,-\frac32. \end{align*}

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$p$ 是 $x^3$ 的系数，所以用 $\binom n3k^3$ ，再代入 $n=16$ 和 $k=-\frac32$ 。

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The coefficient of $x^3$ is

\begin{align*} p=\binom n3k^3. \end{align*}

Using $n=16$ and $k=-\frac32$ ,

\begin{align*} p=&\,\binom{16}{3}\left(-\frac32\right)^3\\ =&\,\frac{16\cdot15\cdot14}{6}\left(-\frac{27}{8}\right)\\ =&\,560\left(-\frac{27}{8}\right)\\ =&\,-1890. \end{align*}