当n≥2时，a_n=1/(2^n-1)=1/(〖(2〗^(n-1)-1)〖(2〗^(n-1)+1))=1/2(1/(2^(n-1)-1)-1/(2^(n-1)+1))s_n=1+1/2 (1-1/3+1/3-1/5+⋯+1/(2^(n-1)-1)-1/(2^(n-1)+1))=1+ 1/2 (1-1/(2^(n-1)+1))=(3∙2^(n-1)+2)...