191问答库 > 已知数列 {an} 的前n项和为Sn=1⼀2n^2+1⼀2n n∈N*

已知数列 {an} 的前n项和为Sn=1⼀2n^2+1⼀2n n∈N*

设bn=1/(（n+2）·an)，且数列{bn}的前n项和为Tn求证：Tn＜3/4

2025-02-23 17:49:54

推荐回答（1个）

回答1：

Sn=1/2n^2+1/2n
a1=S1=1/2+1/2=1
n>=2时
an
=Sn-S(n-1)
=(1/2n^2+1/2n)-(1/2(n-1)^2+1/2(n-1))
=n
∵a1=1满足an=n,n>=2
∴an=n,n∈N*
bn=1/[(n+2)*an]
=1/[(n+2)n]
=1/2*[1/n-1/(n+2)]
∴Tn
=b1+b2+....+bn
=1/2*(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+......
+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2*(1+1/2-1/(n+1)-1/(n+2)]
=1/2*(3/2-(2n+3)/[(n+1)(n+2)])
=3/4-(2n+3)/[2(n+1)(n+2)]
∵n∈N*
∴3/4-(2n+3)/[2(n+1)(n+2)]<3/4
∴Tn<3/4

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