在数列{an}中,a1=1,an+1-an=ln(1+1n),则an=()A.1+n+lnnB.1+nlnnC.1+(n-1)lnnD.1+lnn-数学

题目简介

在数列{an}中,a1=1,an+1-an=ln(1+1n),则an=()A.1+n+lnnB.1+nlnnC.1+(n-1)lnnD.1+lnn-数学

题目详情

在数列{an}中,a1=1,an+1-an=ln(1+
1
n
),则an=(  )
A.1+n+lnnB.1+nlnnC.1+(n-1)lnnD.1+lnn
题型:单选题难度:偏易来源:宜宾一模

答案

∵数列{an}中,a1=1,an+1-an=ln(1+class="stub"1
n
),
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=lnclass="stub"n
n-1
+lnclass="stub"n-1
n-2
+
…+lnclass="stub"2
1
+1
=ln(class="stub"n
n-1
•class="stub"n-1
n-2
•…•class="stub"2
1
)
+1
=lnn+1.
故选D.

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