已知等比數列ＡN的前Ｎ項和為ＳN，若第Ｎ＋１項＝前Ｎ項的和的２倍加上１．求公比
熱心網友
A(n+1)=2Sn+1=S(n+1)-SnS(n+1)=3Sn+1S(n+1)+1/2=3Sn+3/2=3(Sn+1/2)數列{Sn+1/2}是以S1+1/2=A1+1/2為首項,以3為公比的等比數列.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而等比數列前Ｎ項和為ＳN=m*q^n-m所以公比q=3
已知等比數列ＡN的前Ｎ項和為ＳN，若第Ｎ＋１項＝前Ｎ項的和的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)數列{Sn+1/2}是以S1+1/2=A1+1/2為首項,以3為公比的等比數列.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而等比數列前Ｎ項和為ＳN=m*q^n-m所以公比q=3