1、1.在数列an中,a1=1,anan-1=an-1+(-1)n(n2,nN*),则的值是().A.B.C.D.【解析】由已知得an=1+,a2=1+=2,a3=1+=,a4=1+=3,a5=1+=,=.【答案】C2.设数列an中,a1=2,an+1=2an+3,则a4等于().A.30B.35C.37D.40【解析】a2=2a1+3=7,a3=2a2+3=17,a4=2a3+3=37.【答案】C3.已知数列an的通项公式是an=(-1)n(n+1),则a1+a2+a3+a10=.【解析】由an=(-1)n(n+1),得a1+a2+a3+a10=-2+3-4+5-6+7-8+9-10+11=5.
2、【答案】54.已知数列an的通项an=(a,b,c均为正实数),比较an与an+1的大小关系.【解析】an=(a,b,c均为正实数),f(n)=是减函数,an=是增函数,anan+1B.anan+1C.an=an+1D.不能确定【解析】an=,易知an是关于n的增函数,故anan+1.【答案】B7.数列an满足an+1=若a1=,则a20的值为.【解析】逐步计算,可得a1=,a2=-1=,a3=-1=,a4=,a5=-1=,这说明数列an是周期数列,且T=3,所以a36+2=a2=.【答案】8.设函数f(x)=log2x-logx4(0x1),数列an的通项an满足f()=2n(nN*).(1)求数列an的通项公式;(2)证明:数列an是递增数列.【解析】(1)由已知得log2-lo4=2n,即an-=2n,变形整理得-2nan-2=0an=n,又0x1,所以01,故an0,故an+1an.又a2=3+log2=log2log2=,a2a1,an是递增数列.
