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P4 2025 Jan Q3

A Level / Edexcel / P4

IAL 2025 Jan Paper · Question 3

题目

Problem

Given that the binomial expansion, in ascending powers of xx, of

6(4+Ax)−126(4+Ax)^{-\frac{1}{2}}

is

B−14x+Cx2+…B-\frac{1}{4}x+Cx^2+\ldots

(a) find the values of the constants AA, BB and CC.

(5)

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

(1)

For the expansion,

(c) find the coefficient of the term in x3x^3.

(2)
题目中文翻译

已知

6(4+Ax)−126(4+Ax)^{-\frac{1}{2}}

按 xx 的升幂展开的二项式展开式为

B−14x+Cx2+…B-\frac{1}{4}x+Cx^2+\ldots

(a) 求常数 A,B,CA,B,C 的值。

(5)

(b) 写出该展开成立时 xx 的取值范围。

(1)

对于这个展开式,

(c) 求 x3x^3 项的系数。

(2)

解答

(a)

解法一

思路

展开

先提出 44,再展开标准形式 (1+u)−1/2(1+u)^{-1/2},逐项比较常数项、xx 与 x2x^2 的系数。

答题过程

展开

Let u=Ax/4u=Ax/4. Factor out 44:

4+Ax= 4(1+u),6⋅4−1/2= 3.\begin{align*} 4+Ax=&\,4(1+u),\\ 6\cdot4^{-1/2}=&\,3. \end{align*}

Thus 6(4+Ax)−1/2=3(1+u)−1/26(4+Ax)^{-1/2}=3(1+u)^{-1/2}.

The binomial expansion is

(1+u)−1/2=1−12u+38u2+⋯ ,(1+u)^{-1/2}=1-\frac12u+\frac38u^2+\cdots,

because

(−1/2)(−3/2)2=38.\frac{(-1/2)(-3/2)}{2}=\frac38.

Taking u=Ax/4u=Ax/4 gives

 6(4+Ax)−1/2= 3−3A8x+9A2128x2+⋯ .\begin{align*} &\,6(4+Ax)^{-1/2}\\ =&\,3-\frac{3A}{8}x+\frac{9A^2}{128}x^2+\cdots. \end{align*}

Comparing with B−14x+Cx2+⋯B-\frac14x+Cx^2+\cdots, the constant term gives B=3B=3.

The coefficient of xx gives

−3A8= −14,3A= 2,A= 23.\begin{align*} -\frac{3A}{8}=&\,-\frac14,\\ 3A=&\,2,\\ A=&\,\frac23. \end{align*}

The coefficient of x2x^2 gives

C= 9A2128= 9⋅4128⋅9= 132.\begin{align*} C=&\,\frac{9A^2}{128}\\ =&\,\frac{9\cdot4}{128\cdot9}\\ =&\,\frac1{32}. \end{align*}

Therefore A=2/3A=2/3, B=3B=3 and C=1/32C=1/32.

直接对 4+Ax4+Ax 使用二项式公式,也得到相同的常数项与系数 −3A/8-3A/8、9A2/1289A^2/128,不必重复展开。

(b)

解法一

思路

展开

标准展开要求 ∣u∣<1|u|<1,代入 (a) 的 AA 后解出 xx 的范围。

答题过程

展开

The binomial expansion requires ∣Ax/4∣<1|Ax/4|<1. Since A=2/3A=2/3,

∣x6∣< 1,∣x∣< 6.\begin{align*} \bigg|\frac x6\bigg|<&\,1,\\ |x|<&\,6. \end{align*}

Hence the range is

−6<x<6.\boxed{-6<x<6}.

(c)

解法一

思路

展开

计算标准展开的三次项系数,再乘整体因子 33 和代换产生的 (1/6)3(1/6)^3。

答题过程

展开

For (1+u)−1/2(1+u)^{-1/2}, the coefficient of u3u^3 is

(−1/2)(−3/2)(−5/2)3!=−516.\frac{(-1/2)(-3/2)(-5/2)}{3!}=-\frac5{16}.

The expression has an overall factor of 33, so the coefficient of u3u^3 is −15/16-15/16. From (a), u=x/6u=x/6, hence u3=x3/216u^3=x^3/216. Thus the coefficient of x3x^3 is

 −1516⋅216= −51152.\begin{align*} &\,-\frac{15}{16\cdot216}\\ =&\,\boxed{-\frac5{1152}}. \end{align*}