Problem:
Expanding (1+0.2)1000 by the binomial theorem and doing no further manipulation gives
=​(01000​)(0.2)0+(11000​)(0.2)1+(21000​)(0.2)2+⋯+(10001000​)(0.2)1000A0​+A1​+A2​+⋯+A1000​​
where Ak​=(k1000​)(0.2)k for k=0,1,2,…,1000. For which k is Ak​ the largest?
Solution:
For 1≤k≤1000,
Ak−1​Ak​​=(k−1)!(1001−k)!1000!​(0.2)k−1k!(1000−k)!1000!​(0.2)k​=k1001−k​(0.2).
This ratio never equals 1, and exceeds 1 if and only if 1001−k>5k. This last inequality is true just for k≤166. Thus we have
A0​<A1​<…<A166​
while
A166​>A167​>…>A1000​.
Hence Ak​ is largest for k=166​.
The problems on this page are the property of the MAA's American Mathematics Competitions