10.已知SnS_{n}Sn是数列{an}\{a_{n}\}{an}的前nnn项和,若(1−2x)2021=b0+b1x+b2x2+…+b2021x2021(1-2x)^{2021}=b_{0}+b_{1}x+b_{2}x^{2}+\ldots +b_{2021}x^{2021}(1−2x)2021=b0+b1x+b2x2+…+b2021x2021,数列{an}\{a_{n}\}{an}的首项a1=b12+b222+…+b202122021a_{1}=\frac{b_1}{2}+\frac{b_2}{2^2}+\ldots +\frac{b_{2021}}{2^{2021}}a1=2b1+22b2+…+22021b2021,an+1=Sn⋅Sn+1a_{n+1}=S_{n}\centerdot S_{n+1}an+1=Sn⋅Sn+1,则S2021=(S_{2021}=(S2021=( )))