设a_1 = 1,a_n = a_{n-1} / 2.则{a_n}为一等比数列,公比为1/2.显然a_n = 2^{-n+1}.可知,S = 1 + 2^(-1) + 2^(-2) + ...+2^(-2011)=a_1 + a_2 + a_3 + ...+ a_{2012}.即S为等比数列{a_n}的前2012项的和,由等比数列求...