1. Let $f(n)=3 n^{2}-3 n+1$. Find the last four digits of $f(1)+f(2)+\cdots+f(2010)$.
(1 mark)設 $f(n)=3 n^{2}-3 n+1$ 。求 $f(1)+f(2)+\cdots+f(2010)$ 的最後四位數字。
1. 1000
1. Note that $f(n)=n^{3}-(n-1)^{3}$. It follows that
$$
f(1)+f(2)+\cdots+f(2010)=\left(1^{3}-0^{3}\right)+\left(2^{3}-1^{3}\right)+\cdots+\left(2010^{3}-2009^{3}\right)=2010^{3}=201^{3} \times 1000 \text {. }
$$

Since $201^{3}$ has unit digit 1 , it is clear that the answer is 1000 .
Remark. Even without realising $f(n)=n^{3}-(n-1)^{3}$, one could still easily observe that $f(1)+f(2)+\cdots+f(n)=n^{3}$ by computing this sum for some small $n$, which is a natural approach given that the question asks for something related to $f(1)+f(2)+\cdots+f(2010)$.