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CIE 9709 2025 Nov Paper 12 Q2

A Level / CIE / P1

CIE 9709 2025 Nov Paper 12 Paper · Question 2

题目

Problem

Find the term independent of xx in the expansion of (2x2−3x)6\left(2x^2-\frac{3}{x}\right)^6.

[3]
题目中文翻译

求 (2x2−3x)6\left(2x^2-\frac{3}{x}\right)^6 的展开式中不含 xx 的项。

解答

解法一

思路

展开

每选一次 2x22x^2,幂次加 22;每选一次 −3/x-3/x,幂次减 11。先让总幂次等于零,再计算系数。

答题过程

展开

If kk factors supply −3/x-3/x and the other 6−k6-k supply 2x22x^2, the resulting term is

(6k)26−k(−3)kx12−3k.\binom6k 2^{6-k}(-3)^k x^{12-3k}.

The exponent is zero when k=4k=4. The constant term is therefore

(64)22(−3)4=15⋅4⋅81=4860.\binom64 2^2(-3)^4=15\cdot4\cdot81=4860.