题目

Problem

Find, in ascending powers of $x$ up to and including the term in $x^3$ , the binomial expansion of

(1-4x)^{-3}\qquad |x|<\frac{1}{4}

fully simplifying each term.

(4)

题目中文翻译

求

(1-4x)^{-3}\qquad |x|<\frac{1}{4}

按 $x$ 的升幂展开到并包括 $x^3$ 项，并将每一项完全化简。

(4)

解答

展开

使用二项式展开公式。这里可以把 $(1-4x)^{-3}$ 看作 $(1+u)^n$ ，其中 $u=-4x$ 、 $n=-3$ 。题目要求展开到 $x^3$ ，所以保留前四项即可。

展开

For the binomial expansion,

\begin{align*} (1+u)^n =&\,1+nu+\frac{n(n-1)}2u^2\\ &\,\hspace{2pt}+\frac{n(n-1)(n-2)}6u^3+\cdots. \end{align*}

With $n=-3$ , the coefficients are

\frac{(-3)(-4)}2=6,

and

\frac{(-3)(-4)(-5)}6=-10.

Hence

(1+u)^{-3}=1-3u+6u^2-10u^3+\cdots.

Put $u=-4x$ :

\begin{align*} (1-4x)^{-3} =&\,1-3(-4x)+6(-4x)^2\\ &\,\hspace{2pt}-10(-4x)^3+\cdots. \end{align*}

The three non-constant terms simplify to

-3(-4x)=12x,

6(-4x)^2=6(16x^2)=96x^2,

and

-10(-4x)^3=-10(-64x^3)=640x^3.

Therefore

\begin{align*} (1-4x)^{-3} =&\,1+12x+96x^2\\ &\,\hspace{2pt}+640x^3+\cdots, \end{align*}

for $|x|<\frac14$ .