题目

Problem

Given that the binomial expansion, in ascending powers of $x$ , of

6(4+Ax)^{-\frac{1}{2}}

B-\frac{1}{4}x+Cx^2+\ldots

(a) find the values of the constants $A$ , $B$ and $C$ .

(5)

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

(1)

For the expansion,

(2)

题目中文翻译

已知

6(4+Ax)^{-\frac{1}{2}}

按 $x$ 的升幂展开的二项式展开式为

B-\frac{1}{4}x+Cx^2+\ldots

(a) 求常数 $A,B,C$ 的值。

(b) 写出该展开成立时 $x$ 的取值范围。

对于这个展开式，

解答

Rewrite the expression in a form suitable for the binomial expansion:

\begin{aligned} 6(4+Ax)^{-\frac12} &=6\left(4\left(1+\frac{A}{4}x\right)\right)^{-\frac12} \\ &=6\cdot4^{-\frac12}\left(1+\frac{A}{4}x\right)^{-\frac12} \\ &=3\left(1+\frac{A}{4}x\right)^{-\frac12} \end{aligned}

Using

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

with

n=-\frac12,\qquad u=\frac{A}{4}x

we get

\begin{aligned} 3\left(1+\frac{A}{4}x\right)^{-\frac12} &=3\left(1-\frac12\left(\frac{A}{4}x\right) +\frac{\left(-\frac12\right)\left(-\frac32\right)}{2} \left(\frac{A}{4}x\right)^2+\cdots\right) \\ &=3-\frac{3A}{8}x+\frac{9A^2}{128}x^2+\cdots \end{aligned}

Compare this with

B-\frac14x+Cx^2+\cdots

First,

B=3

For the coefficient of $x$ ,

-\frac{3A}{8}=-\frac14

\frac{3A}{8}=\frac14

and hence

A=\frac23

For the coefficient of $x^2$ ,

C=\frac{9A^2}{128}

Substitute $A=\dfrac23$ :

C=\frac{9}{128}\left(\frac23\right)^2 =\frac{9}{128}\cdot\frac49 =\frac1{32}

Therefore

\boxed{A=\frac23,\qquad B=3,\qquad C=\frac1{32}}

For the binomial expansion to be valid, we need

\left|\frac{A}{4}x\right|<1

Since $A=\dfrac23$ ,

\left|\frac{1}{6}x\right|<1

|x|<6

Hence the range of validity is

\boxed{-6<x<6}

The coefficient of the $x^3$ term is

3\cdot \frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{3!} \left(\frac{A}{4}\right)^3

Using $A=\dfrac23$ ,

\frac{A}{4}=\frac16

Therefore the coefficient is

\begin{aligned} 3\cdot \frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6} \left(\frac16\right)^3 &= 3\cdot\left(-\frac{15}{8}\right)\cdot\frac16\cdot\frac1{216} \\ &=-\frac{45}{10368} \\ &=-\frac5{1152} \end{aligned}

Hence the coefficient of the term in $x^3$ is

\boxed{-\frac5{1152}}