已知数列{an}的a1=2,设其前n项和为Sn,且对任意的n∈N+,n≥2,an总是3Sn-4和2-
Sn-1的等差中项,则下列各式成立的是( )
①Sn•Sn+2>S2n+1;
②Sn•Sn+2<S2n+1;
③Sn+Sn+2<2Sn+1
④Sn+Sn+2>2Sn+1.
| 5 |
| 2 |
①Sn•Sn+2>S2n+1;
②Sn•Sn+2<S2n+1;
③Sn+Sn+2<2Sn+1
④Sn+Sn+2>2Sn+1.
| A、①② | B、②③ | C、③④ | D、①④ |
| 5 |
| 2 |
| A、①② | B、②③ | C、③④ | D、①④ |