Problem:
Let xn​=sin2(n∘). What is the mean of x1​,x2​,x3​,…,x90​ ?
Answer Choices:
A. 4511​
B. 4522​
C. 18089​
D. 21​
E. 18091​
Solution:
The required mean is
901​n=1∑90​sin2(n∘)
Group the summands into 44 pairs plus two additional terms as follows:
sin2(1∘)+sin2(89∘)sin2(2∘)+sin2(88∘)sin2(3∘)+sin2(87∘)⋮sin2(44∘)+sin2(46∘)sin2(45∘)sin2(90∘)​=sin2(1∘)+cos2(1∘)=1=sin2(2∘)+cos2(2∘)=1=sin2(3∘)+cos2(3∘)=1=sin2(44∘)+cos2(44∘)=1=21​=1​
This gives a sum of 4521​=291​, so the mean is 901​⋅291​=(E)18091​​.
The problems on this page are the property of the MAA's American Mathematics Competitions