题目
Problem
In the expansion of
(px+3)5−(x3+xp)4,
the coefficient of x4 is 216.
Find the value of the positive constant p.
[5]
题目中文翻译
在
(px+3)5−(x3+xp)4
的展开式中,x4 的系数为 216。
求正常数 p 的值。
解答
解法一
思路
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分别找出两个展开式的 x4 项。第二个展开式里,每把一个 x3 换成 p/x,x 的幂次会减少 4。
答题过程
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In (px+3)5, the x4 term is (45)(px)4⋅3=15p4x4. In the second expansion, choosing k copies of p/x gives exponent 3(4−k)−k=12−4k. This equals 4 when k=2, so its x4 term is (24)(x3)2(p/x)2=6p2x4.
Subtracting the coefficients gives 15p4−6p2=216. Dividing by 3 and writing u=p2 gives
5u2−2u−72=(u−4)(5u+18)=0.
Since p>0, we require u>0. Thus u=4 and p=2.