题目内容
(2011•南充一模)在数列{an}中,a1=2,an+1=an+ln(1+
),则an=( )
1 |
n |
分析:利用“累加求和”和对数的运算法则即可得出.
解答:解:∵在数列{an}中,a1=2,an+1=an+ln(1+
),∴an+1-an=ln
.
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=ln
+ln
+…+ln
+2
=ln(
•
•…•
)+2
=lnn+2.
故选A.
1 |
n |
n+1 |
n |
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=ln
n |
n-1 |
n-1 |
n-2 |
2 |
1 |
=ln(
n |
n-1 |
n-1 |
n-2 |
2 |
1 |
=lnn+2.
故选A.
点评:熟练掌握“累加求和”和对数的运算法则是解题的关键.
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