在数列{an}中,a1=2,an+1=an+ln
,则an=( )
A.2+lnn B.2+(n-1)lnn
C.2+nln
n D.1+n+lnn
在数列{an}中,a1=2,an+1=an+ln,则an=( )
A.2+lnn B.2+(n-1)lnn
C.2+nlnn D.1+n+lnn
A.由已知得
an+1-an=ln
=ln(n+
1)-lnn.
于是an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)
=2+(ln2-ln1)+(ln3-ln2)+…+[lnn-ln(n-1)]=2+lnn.