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−1Ak=(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