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
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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
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
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
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
列：在等差数列{an}中,a1<0,Sn为前n项和,且S3=S16,则Sn取得最小值的n值为
在等差数列{an}中,a1<0,Sn为前n项和,且S3=S16,则Sn取得最小值的n值为?
从S3=S16可推出a10=0
因为an是等差数列,a1〈0,a10=0,可知d≠0
所以Sn取得最小值的n值为10
点(Sn+1,Sn)在直线x/(n+1)-y/n=1上 可知数列Sn/n成等差
先求数列Sn/n的通项，然后可求得Sn的通项，再根据Sn-Sn-1=an求得an
做到第一问再看看第二问能不能迎刃而解