若数列{an}是正项数列,且a1+a2+…+an=n2+3n(n∈N*),则a12+a23+…+ann+1=______.-数学

题目简介

若数列{an}是正项数列,且a1+a2+…+an=n2+3n(n∈N*),则a12+a23+…+ann+1=______.-数学

题目详情

若数列{an}是正项数列,且
a1
+
a2
+…+
an
=n2+3n(n∈N*),则
a1
2
+
a2
3
+…+
an
n+1
=______.
题型:填空题难度:中档来源:不详

答案

令n=1,得
a1
=4,∴a1=16.
当n≥2时,
a1
+
a2
+…+
an-1
=(n-1)2+3(n-1).
与已知式相减,得
an
=(n2+3n)-(n-1)2-3(n-1)=2n+2,
∴an=4(n+1)2,n=1时,a1适合an.
∴an=4(n+1)2,
an
n+1
=4n+4,
a1
2
+
a2
3
++
an
n+1
=
n(8+4n+4)
2
=2n2+6n.
故答案为2n2+6n

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