Given that $$1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n(n+1)(2n+1)}{6}$$, then $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=$$． 
Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas->1²+2²+3²+......+n²=1/6n(n+1)(n+2)
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 $$\text{D}$$. 