向量a=(x^2+1,-x)b=(1,2√n^2+1)(n为正整数)函数f(x)=a·b 设f(x)在(0,+∞)上取最小值时的自变量x取值为an
问题描述:
向量a=(x^2+1,-x)b=(1,2√n^2+1)(n为正整数)函数f(x)=a·b 设f(x)在(0,+∞)上取最小值时的自变量x取值为an
1.求数列an的通项公式
2.已知数列bn,对任意正整数n都有bn·(4(an)^2-5)=1成立 设Sn为数列bn的前n项和求limSn
答
(1)f(x)=x^2+1-2x√n^2+1f'(x)=2x-2√n^2+1令f'(x)=0得x=√n^2+1 (没有学过导数的可以用判别式来做)即an=√n^2+1(2)将an代入得bn=1/(4n^2-1)=1/(2n+1)(2n-1)=1/2[1/(2n-1)-1/(2n+1)] (以后看见平方差的倒数时要...