Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 June Paper 13 Q3

A Level / CIE / P1

CIE 9709 2023 June Paper 13 Paper · Question 3

题目

Problem

(a) Give the complete expansion of (x+2x)5\bigg(x + \frac{2}{x}\bigg)^5.

[2]

(b) In the expansion of (a+bx2)(x+2x)5(a + bx^2)\bigg(x + \frac{2}{x}\bigg)^5, the coefficient of xx is zero and the coefficient of 1x\frac{1}{x} is 8080.

Find the values of the constants aa and bb.

[4]
题目中文翻译

(a) 写出 (x+2x)5\bigg(x + \frac{2}{x}\bigg)^5 的完整展开式。

[2]

(b) 在 (a+bx2)(x+2x)5(a + bx^2)\bigg(x + \frac{2}{x}\bigg)^5 的展开式中,xx 的系数为零,1x\frac{1}{x} 的系数为 8080。

求常数 aa 和 bb 的值。

[4]

解答

(a)

解法一

思路

展开

每多选一个 2/x2/x,幂次减少 22。按二项式定理依次写出六项。

答题过程

展开

Using the general term (5j)2jx5−2j\binom5j2^j x^{5-2j} for j=0,…,5j=0,\ldots,5, the expansion is

 x5+10x3+40x+80x−1 +80x−3+32x−5.\begin{align*} &\,x^5+10x^3+40x+80x^{-1}\\ &\,+80x^{-3}+32x^{-5}. \end{align*}

(b)

解法一

思路

展开

乘上 bx2bx^2 会把幂次提高 22。分别找出产生 xx 和 x−1x^{-1} 的两组项。

答题过程

展开

The coefficient of xx is 40a+80b40a+80b, and that of x−1x^{-1} is 80a+80b80a+80b. Hence

40a+80b=0,80a+80b=80.40a+80b=0,\qquad80a+80b=80.

Subtracting the equations gives 40a=8040a=80, so a=2a=2 and b=−1b=-1.