已知等比数列ＡN的前Ｎ项和为ＳN，若第Ｎ＋１项＝前Ｎ项的和的２倍加上１．求公比
A(n+1)=2Sn+1=S(n+1)-SnS(n+1)=3Sn+1S(n+1)+1/2=3Sn+3/2=3(Sn+1/2)数列{Sn+1/2}是以S1+1/2=A1+1/2为首项,以3为公比的等比数列.Sn+1/2=(A1+1/2)*3^(n-1)Sn=(A1+1/2)*3^(n-1)-1/2=[(A1+1/2)/3]*3^n-1/2而等比数列前Ｎ项和为ＳN=m*q^n-m所以公比q=3
