Problem:
Let f(x)=(x2+3x+2)cos(πx). Find the sum of all positive integers n for which
∣∣∣∣∣∣k=1∑nlog10f(k)∣∣∣∣∣∣=1
Solution:
Note that
f(k)=[(k+1)(k+2)](−1)k={(k+1)(k+2)(k+1)(k+2)1 if k is even if k is odd
Therefore
k=1∑nlog10f(k)=log10(k=1∏nf(k))=⎩⎪⎨⎪⎧log10(2⋅3⋅4⋯(n+1)3⋅4⋅5⋯(n+2))=log10(2n+2)log10(2⋅3⋅4⋯(n+2)3⋅4⋅5⋯(n+1))=log10(2(n+2)1)=−log10(2n+4) if n is even if n is odd.
For ∣∑k=1nlog10f(k)∣ to be 1, either 2n+2=10 with n even or 2n+4=10 with n odd, so n=18 or n=3. Thus the requested sum is 18+3=21.
The problems on this page are the property of the MAA's American Mathematics Competitions