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 $a_n < \frac{1}{2018}.$
+118608 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.
