2009年AMC10竞赛B第14题
$$2009-AMC10B-14$$  On Monday, Millie puts a quart of seeds, $$25 \%$$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $$25 \%$$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?  星期一，米莉将一夸脱种子（其中 25\% 是小米）放入喂鸟器中。 在连续的每一天，她添加另一夸脱相同的种子混合物，但没有去除任何剩余的种子。 这些鸟每天只吃喂食器中 25\% 的小米，但它们吃掉所有其他种子。 哪一天，米莉刚放完种子，鸟儿会发现喂食器里一半以上的种子是小米？
美国AMC10/12->Knowledge Point->Algebra->Sequence->Sum of the First n Terms for Geometric Sequences 课内体系->知识点->数列->数列的实际应用
On Monday, day $$1$$, the birds find $$\frac14$$ quart of millet in the feeder. On Tuesday they find $$\frac{1}{4}+ \frac{3}{4}\cdot \frac{1}{4}$$ quarts of millet. On Wednesday, day $$3$$, they find $$\frac{1}{4}+ \frac{3}{4}\cdot \frac{1}{4}+\left(\frac{3}{4}\right)^{2}\cdot \frac{1}{4}$$ quarts of millet. The number of quarts of millet they find on day $$n$$ is $$\frac{1}{4}+ \frac{3}{4}\cdot \frac{1}{4}+\left(\frac{3}{4}\right)^{2}\cdot \frac{1}{4}+ \cdots +\left(\dfrac{3}{4}\right)^{n-1}\cdot \frac{1}{4}= \dfrac{\left(\dfrac{1}{4}\right)\left(1-\left(\dfrac{3}{4}\right)^{n}\right)}{1- \dfrac{3}{4}}=1-\left(\frac{3}{4}\right)^{n}$$. The birds always find $$\frac34$$ quart of other seeds, so more than half the seeds are millet if $$1-\left(\frac{3}{4}\right)^{n}\textgreater\frac34$$ , that is, when $$\left(\frac{3}{4}\right)^{n}\textless{} \frac{1}{4}$$. Because $$\left(\frac{3}{4}\right)^{4}= \frac{81}{256}\textgreater{} \frac{1}{4}$$ and $$\left(\frac{3}{4}\right)^{5}= \frac{243}{1024}\textless{} \frac{1}{4}$$, this will first occur on day $$5$$ which is $$\boxed{\rm Friday}$$. The answer is $$\rm(D)$$. 