Problem:
For all integers n greater than 1 , define an=logn20021. Let b=a2+a3+a4+a5 and c=a10+a11+a12+a13+a14. Then b−c equals
Answer Choices:
A. −2
B. −1
C. 20021
D. 10011
E. 21
Solution:
We have an=logn20021=log2002n, so
b−c=(log20022+log20023+log20024+log20025)−(log200210+log200211+log200212+log200213+log200214)=log200210⋅11⋅12⋅13⋅142⋅3⋅4⋅5=log200211⋅13⋅141=log200220021=−1
The problems on this page are the property of the MAA's American Mathematics Competitions