αn是正项数列
发布网友
发布时间：2022-04-29 18:59
共1个回答
热心网友
时间：2022-06-20 02:57
√a1 + √a2+ . + √a(n-1)=(n-1)^2+3(n-1)=n^2-2n+1+3n-3=n^2+n-2.2
1式-2式得
√an=n^2+3n-(n^2+n-2)
√an=n^2+3n-n^2-n+2
√an=2n+2
an=4(n+1)^2
(a1)/2+(a2)/3+.(an)/(n+1)
=4*2^2/2+4*3^2/3+.+4(n+1)^2/(n+1)
=4[2^2/2+3^2/3+.+(n+1)^2/(n+1)]
=4*(2+3+4+.+n+1)
=4*(2+n+1)*n/2
=2n(n+3)
=2n^2+6n
