5、an+1=an+ln,则a5=( D )A.1+ln2 B.2+ln3 C.3+ln5 D.2+ln5解析因为an+1-an=ln=ln=ln(n+1)-lnn,所以a5-a1=(a5-a4)+(a4-a3)+(a3-a2)+(a2-a1)=(ln5-ln4)+(ln4-ln3)+(ln3-ln2)+(ln2-ln1)=ln5-ln1=ln5,所以a5=a1+ln5=2+ln5,故选D.3.若数列的通项公式为an=2n+2n-1,则数列的前n项和为( C )A.2n+n2-1 B.2n+1+n2-1C.2n+1+n2-2
6、 D.2n+n-2解析Sn=a1+a2+a3+…+an=(21+2×1-1)+(22+2×2-1)+(23+2×3-1)+…+(2n+2n-1)=(2+22+…+2n)+2(1+2+3+…+n)-n=+2×-n=2(2n-1)+n2+n-n=2n+1+n2-2.4.若数列的通项公式是an=(-1)n(3n-2),则a1+a2+a3+…+a10=( A )A.15 B.12 C.-12 D.-15解析∵an=(-1)n(3n-2),∴a1+a2+a3+…+a10=-1+4-7+10-13+16-19+22-25+28=(-