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Let $a_1,$ $a_2,$ $\dots$ be a sequence of real numbers such that for all positive integers $n,$ \[\sum_{k = 1}^n a_k \left( \frac{k}{n} \right)^2 = 1.\]Find the smallest $n$ such that $a_n < \frac{1}{2018}.$

 Aug 8, 2021
 #1
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The smallest n that works is 2019.

 Aug 8, 2021
 #2
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I don't get this question.  It does not make sense to me. 

 

If

\(\displaystyle \sum_{k = 1}^n a_k \left( \frac{k}{n} \right)^2 = 1.\\ then\\ \displaystyle \sum_{k = 1}^1 a_1 \left( \frac{1}{1} \right)^2 = 1.\\ so\\a_1=1 \)

 

Since the sum of any number of terms must equal 1    

\(a_k=0\quad \text{Where k is an integer greater than 1}\)

 

Obviously, I am not understanding something.

 Aug 8, 2021

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