Directly substitute $$n$$ with $$18$$ into the formula to get $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=\frac{18\times 19\times 37}{6}=2109$$. So the answer is $$\text{D}$$. 
Given that $$1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n(n+1)(2n+1)}{6}$$, then $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=$$． 
  - $$2016$$ 
  - $$2107$$ 
  - $$2018$$ 
  - $$2109$$ 
  - $$2020$$ 