已知数列\(\{a_{n}\}\)满足\(a_{n}+1=a_{n+1}\)且\(a_{2}+a_{4}=6\)，当\(1\leqslant n\leqslant 2020\)，\(n\in N^{*}\)时，记\(S_{n}=a_{1}+a_{2}+⋯+a_{n}\)，则\(S_{1}+S_{2}+⋯+S_{20}=\)______\(.(\)备用公式\(1^{2}+2^{2}+⋯+n^{2}=\dfrac{n(n+1)(2n+1)}{6})\)