题目

Problem

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

\frac{8}{(2-5x)^2}

writing each term in simplest form.

(4)

(b) Find the range of values of $x$ for which this expansion is valid.

(1)

题目中文翻译

(a) 求

\frac{8}{(2-5x)^2}

按 $x$ 的升幂展开时的前四项，并将每一项写成最简形式。

(b) 求该展开式成立时 $x$ 的取值范围。

解答

展开

先从分母中提出 $2^2$ ，把原式改写成 $2(1-\frac{5x}{2})^{-2}$ 。再使用二项式展开

(1+u)^n =1+nu+\frac{n(n-1)}{2!}u^2 +\frac{n(n-1)(n-2)}{3!}u^3+\cdots

并取 $n=-2,\ u=-\frac{5x}{2}$ ，即可得到按 $x$ 升幂排列的前四项。

展开

First rewrite the expression as

\frac{8}{(2-5x)^2} =2\left(1-\frac{5x}{2}\right)^{-2}.

Using

(1+u)^{-2} =1-2u+3u^2-4u^3+\cdots

with $u=-\frac{5x}{2}$ gives

\begin{align*} \frac{8}{(2-5x)^2} &=2\left[ 1-2\left(-\frac{5x}{2}\right) +3\left(-\frac{5x}{2}\right)^2 -4\left(-\frac{5x}{2}\right)^3 +\cdots \right]\\ &=2\left( 1+5x+\frac{75}{4}x^2+\frac{125}{2}x^3+\cdots \right). \end{align*}

Therefore, the first four terms are

\boxed{2+10x+\frac{75}{2}x^2+125x^3}.

展开

负整数指数的二项式展开要求括号内的变量部分绝对值小于 $1$ 。这里变量部分是 $-\frac{5x}{2}$ ，解对应的绝对值不等式即可。

展开

The expansion is valid when

\left|-\frac{5x}{2}\right|<1.

Hence

|x|<\frac{2}{5}.

Therefore, the required range is

\boxed{-\frac{2}{5}<x<\frac{2}{5}}.