> 数学 >
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
=______.
人气:132 ℃ 时间:2019-10-26 17:58:03
解答
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,∴ann+1=4n+4,∴a12+a23++ann+1...
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