一到数列题目 高手进

来源：百度知道 编辑：UC知道 时间：2024/05/08 12:24:16
数列前n项和为Sn a1=2 点(Sn+1,Sn)在直线x/(n+1)-y/n=1上 n是正整数
1.求an通项
2.设Tn=(Sn/S(n+1))+(S(n+1))/Sn-2 证明4/3≤T1+T2+T3+…+Tn<3

S(n+1)/(n+1) - S(n)/n = 1,

S(1) = a(1) = 2.

设b(n) = S(n)/n, n = 1,2,...
则 b(n+1) - b(n) = 1.

{b(n)}是首项为b(1) = S(1)/1 = 2，公差为1的等差数列。
b(n) = 2 + (n-1) = n+1.
S(n)/n = n+1,

S(n) = n(n+1). n = 1,2,...
a(1) = 2,
n >= 2时，
a(n) = S(n) - S(n-1) = n(n+1) - (n-1)n = 2n,

a(n) = 2n, n = 1,2,...

T(n) = S(n)/S(n+1) + S(n+1)/S(n) - 2 = n(n+1)/[(n+1)(n+2)] + (n+1)(n+2)/[n(n+1)] - 2

= n/(n+2) + (n+2)/n - 2

= 1 - 2/(n+2) + 1 + 2/n - 2

= 2/n - 2/(n+2)

G(n) = T(1) + T(2) + T(3) + ...+ T(n-2) + T(n-1) + T(n)

= 2/1 - 2/3 + 2/2 - 2/4 + 2/3 - 2/5 + ... + 2/(n-2) - 2/n + 2/(n-1) - 2/(n+1) + 2/n - 2/(n+2)

= 2/1 + 2/2 - 2/(n+1) - 2/(n+2)

= 3 - 2[1/(n+1) + 1/(n+2)]

所以
G(n) = 3 - 2[1/(n+1) + 1/(n+2)]< 3,

G(n) = 3 - 2[1/(n+1) + 1/(n+2)] >= 3 - 2[1/2 + 1/3] = 3 - 1 - 2/3 = 4/3.

综合，有
4/3 <