Question

在數列{a_n}中,a₁=2,a_n+1=a_n+ln(1+),則a_n=(　　)

A．2+lnn

B．2+(n-1)lnn

C．2+nlnn

D．1+n+lnn

Answer 1

A

【思路點撥】根據遞推式采用“疊加”方法求解.
解:∵a_n+1=a_n+ln(1+)=a_n+ln=a_n+ln(n+1)-lnn,
∴a₂=a₁+ln2,a₃=a₂+ln3-ln2,…,a_n=a_n-1+lnn-ln(n-1),
將上面n-1個式子左右兩邊分別相加得a_n=a₁+ln2+(ln3-ln2)+(ln4-ln3)+…+[lnn-ln(n-1)]=a₁+lnn=2+lnn.