Question

已知数列{an}前n项和为Sn，且n∈正整数都有Sn=an+n²/2

（1）求a1、a2及an
(2)若数列{bn}满足b1=1,当n≥2时bn=an²（1/a1+1/a2+..........+1/an²-1)
证明：n≥2时，bn+1/(n+1)²-bn/n²=1/n² 前面的bn+1是第n+1项
（3）在（2)的条件下比较（1+1/b1)(1+1/b2)(1+1/b3).......(1+1/bn)与4的关系
求高人相助，给出第三问的步骤和方法，急求 希望好心人帮一下

Answer 1

1,题目是不是S_n=(a_n+n^2)/2?否则，题目有问题
若S_n=(a_n+n^2)/2，则a_1=1
S_n-S_(n-1)=a_n=[a_n-a_(n-1)+n^2-(n-1)^2)]/2==[a_n-a_(n-1)+2n-1)]/2
得a_n+a_(n-1)=2n-1,得a_2=2,.....a_n=n
(2)b_n=an²（1/a1+1/a2+..........+1/an²-1)中an²是什么？是第n²项吗 若是
则b_n=n^2(1+1/2+.....+1/n^2-1)
则bn+1/(n+1)²-bn/n²=1/(n+1)²

Answer 2

(1)
a(1)=s(1)=[a(1)+1]/2, a(1)=s(1)=1,
a(n+1)=s(n+1)-s(n)=[a(n+1)+(n+1)^2]/2 - [a(n)+n^2]/2 = [a(n+1)-a(n)+2n+1]/2,
a(n+1)=-a(n)+2n+1=-a(n)+n+n+1,
a(n+1)-(n+1)=-[a(n)-n],
{a(n)-n}是首项为a(1)-1=0,公比为(-1)的等比数列.
a(n)-n=0
a(n) = n,
a(2)= 2.
s(2)=a(1)+a(2)=1+2=3, [a(2)+2^2]/2=[2+4]/2=3=s(2).
(2)
b(1)=1,
b(n+1)=[a(n+1)]^2{1/[a(1)]^2 + 1/[a(2)]^2 + ... + 1/[a(n)]^2}=(n+1)^2{1/1 + 1/2^2 + ... + 1/n^2},
b(n+2)=(n+2)^2{1/1 + 1/2^2 + ... + 1/n^2 + 1/(n+1)^2} = (n+2)^2{1/1 + 1/2^2 + ... +1/n^2} + (n+2)^2/(n+1)^2
=(n+2)^2/(n+1)^2*b(n+1) + (n+2)^2/(n+1)^2,
b(n+2)/(n+2)^2 - b(n+1)/(n+1)^2 = 1/(n+1)^2.
n>=2时,
b(n+1)/(n+1)^2 - b(n)/n^2 = 1/n^2
(3)
b(1)=1,
b(n+1)=(n+1)^2{1/1 + 1/2^2 + ... + 1/n^2},
b(n+2)/(n+2)^2 - b(n+1)/(n+1)^2 = 1/(n+1)^2,
b(n+2) = (n+2)^2[b(n+1)+1]/(n+1)^2,

比较烦,...

Answer 3

S1=a1+1/2=a1.....这不和逻辑啊。。。Sn=an+n²/2