二次函数y=n(n+1)x2-(2n+1)x+1,当n依次取1,2,3,4,…,n,…时,图象在x轴上截得的线

你这个题吧,题目给的不完整二次函数y=n(n+1)x2-(2n+1)x+1由于△=(2n+1)^2-4n(n+1)=(4n^2+4n+1)-(4n^2+4n)=1>0∴方程y=n(n+1)x2-(2n+1)x+1=0总是有两个不同的根即二次曲线与x轴总是有两个交点,图像在x轴上截得的线段为an由y=n(n+1)x2-(2n+1)x+1=(nx-1)[(n+1)x-1]=0解得两个根为x1=1/n, x2=1/(n+1)∴an=x1-x2=1/n-1/(n+1)=1/[n(n+1)]这个截得的线段是随n的取值的增大而减小的求lim,猜测有两种可能：一是求线段an的极限,二是求所有线段之和Sn的极限若是前者,则有liman=lim{1/[n(n+1)]}=0 (n->+∞)若是后者,则有limSn=lim(a1+a2+...+an)=lim{(1/1-1/2)+(1/2-1/3)+...+[1/n-1/(n+1)]}=lim(1-1/(n+1)) (n->+∞)=1

y=n(n+1)x^2-(2n+1)x++1
y=(nx-1)[(n+1)x-1]
x1=1/n x2=1/(n+1)
x轴上截得的线
an=1/n-1/(n+1)
Sn=1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)