设等比数列{a[n]}的公比为q则S[n]=a[1](1-qⁿ)/(1-q)=2(1-qⁿ)/(1-q)则S[n]+1=2(1-qⁿ)/(1-q)+1S[1]+1=2(1-q)/(1-q)+1=2+1=3S[2]+1=2(1-q²)/(1-q)+1=2(1+q)+1=3+2qS[3]+1=2(1-q³)/(1-q)+1=...