1.在数列{an}中,若a1=2,an=(n≥2,n∈N*),则a8=( )
A.-1B.1C.D.2
[解析]因为a1=2,an=(n≥2,n∈N*),所以a2==-1,a3==,a4==2,所以{an}是周期数列,周期是3,所以a8=a2=-1.
[答案]A
2.已知数列{an}满足a1=1,an+1=则其前6项之和为( )
A.16B.20C.33D.120
[解析]a2=2a1=2,a3=a2+1=3,a4=2a3=6,a5=a4+1=7,a6=2a5=14,所以S6=1+2+3+6+7+14=33.
[答案]C
3.已知数列{an}的前n项和Sn满足:Sn+Sm=Sn+m,且a1=1,那么a10=( )
A.1B.9C.10D.55