作业帮已知数列{an}的前n项和为sn,点（n,sn（n属于n+）在函数fx=1/2（x^2）+1/2x 1）求an通项公式 2）设数列{1/（anan+2}的前n项和为tn,不等式tn>1/3log a （1-a）对任意正整数n恒成立,求实数a取值范围
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/05 21:40:44
已知数列{an}的前n项和为sn,点（n,sn（n属于n+）在函数fx=1/2（x^2）+1/2x 1）求an通项公式 2）设数列{1/（anan+2}的前n项和为tn,不等式tn>1/3log a （1-a）对任意正整数n恒成立,求实数a取值范围
已知数列{an}的前n项和为sn,点（n,sn（n属于n+）在函数fx=1/2（x^2）+1/2x 1）求an通项公式 2）设数列{1/（anan+2}的前n项和为tn,不等式tn>1/3log a （1-a）对任意正整数n恒成立,求实数a取值范围
已知数列{an}的前n项和为sn,点（n,sn（n属于n+）在函数fx=1/2（x^2）+1/2x 1）求an通项公式 2）设数列{1/（anan+2}的前n项和为tn,不等式tn>1/3log a （1-a）对任意正整数n恒成立,求实数a取值范围
s(n) = n^2/2 + n/2,
a(1)=s(1)=1/2+1/2=1,
s(n+1)=(n+1)^2/2 + (n+1)/2,
a(n+1)= s(n+1) -s(n) = (2n+1)/2 + 1/2 = n+1,
a(n) = n.
b(n) = 1/[a(n)a(n+2)] = 1/[n(n+2)] = (1/2)[1/n - 1/(n+2)] = (1/2)[1/n + 1/(n+1)] - (1/2)[1/(n+1) + 1/(n+2)],
t(n) = b(1)+b(2)+...+b(n-1)+b(n) = (1/2)[1/1 + 1/2] - (1/2)[1/2+1/3] + (1/2)[1/2 + 1/3]-(1/2)[1/3+1/4] + ... + (1/2)[1/(n-1)+1/n]-(1/2)[1/n+1/(n+1)] + (1/2)[1/n+1/(n+1)] - (1/2)[1/(n+1)+1/(n+2)]
=(1/2)[1/1 + 1/2] - (1/2)[1/(n+1)+1/(n+2)]
= 3/4 - (1/2)[1/(n+1) + 1/(n+2)]
{t(n)}单调递增,1/3 = t(1) (1/3)log_{a}{1-a}, 0