由通项可得1/(d2-d1)+1/(d3-d2)+...+1/(d(n+1)-dn)=1/(2^3-2^1)+1/(2^5-2^3)+1/(2^7-2^5)+.+1/[2^(2n+1)-2^(2n-1)]=1/3*[1/2+1/2^3+1/2^5+.+1/2^(2n-1)]=1/3*1/2*[1-(1/2)^(2n)]/(1-1/4)=2/9*[1-(1/2)^(2n)]