再数列{An}中，A1=2,An+1=An+In(1+1/n)则An=2+Inn

A1=2 An+1=An+ln(1+1/n) 则
A2=2+ln(1+1/1)
A3=2+ln(1+1/1)+ln(1+1/2)
…………
可知An=2+ln(1+1/1)+ln(1+1/2)+……ln(1+1/n-1)
=2+ln(2*3/2*4/3*……*n/n-1)
=2+lnn

解：（用数学归纳法）
∵a1=2，a（n+1）= an+In(1+1/n)
∴当n=1时，
a2= a1 +In(1+1/1)=2+ In2
当n=1时，
a3= a2 +In(1+1/2)=（2+ In2）+ In(1+1/2)
=2+ In2 +In(3/2)= 2+ In3
设当n=k-1时有ak=2+ Ink成立
则当n=k时，a（k+1）= ak+In(1+1/k)
=（2+ Ink）+ In[(k+1)/k]
=2+ Ink+ In(k+1)- Ink
=2+In(k+1)
∴当n=k+1时，得An=2+Inn