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CIE 9709 2025 June Paper 12 Q3

A Level / CIE / P1

CIE 9709 2025 June Paper 12 Paper · Question 3

题目

Problem

The coefficient of x7x^7 in the expansion of (px2+4px)5\left(px^2 + \frac{4}{p}x\right)^5 is 1280.

Find the value of the constant pp.

[4]
题目中文翻译

在 (px2+4px)5\left(px^2 + \frac{4}{p}x\right)^5 的展开式中,x7x^7 的系数为 1280。

求常数 pp 的值。

[4]

解答

解法一

思路

展开

若五个因子全选一次项,幂次为 55。每换一个二次项,幂次增加 11,所以只需选两个二次项。

答题过程

展开

The term in x7x^7 comes from two factors px2px^2 and three factors 4x/p4x/p. Its coefficient is

(52)p2(4p)3=640p.\binom52 p^2\Big(\frac4p\Big)^3=\frac{640}{p}.

Since p≠0p\ne0 and this coefficient equals 12801280, we obtain p=12p=\frac12.