已知f(x)=x^2-2(n+1)x+n^2+5n-7(1)设f(x)的图像的顶点的横坐标构成数列{an}，求证{an}为等差数列。(2) 设f(x)的图像的顶点到x轴的距离构成数列{bn}，求{bn}的前n项和Sn
热心网友
解:(1)f(x)=x^-2(n+1)x+n^+5n-7=x^-2(n+1)x+(n^+2n+1)+3n-8=(x-n-1)^+3n-8∴顶点的横坐标:an=n+1。n≥ -an-1=(n+1)-n=1∴{an}为等差数列。(2) 顶点的纵坐标:3n-8f(x)的图像的顶点到x轴的距离∴bn=|3n-8|当n≤2,3n-8＜0,当n≥3,3n-8＞0①当n≤2,即:Sn =|3×1-8|+|3×2-8|=(13-3n)n/2②当n≥3,即:Sn =|3×1-8|+|3×2-8|+|3×3-8|+|3×4-8|+…+|3n-8|=-(3×1-8)-(3×2-8)+(3×3-8)+(3×4-8)+…+(3n-8)=[-2(3×1-8)-2(3×2-8)]+(3×1-8)+(3×2-8)+(3×3-8)+(3×4-8)+…+(3n-8)=[-2(3×1-8)-2(3×2-8)]+(3n-13)n/2=14+(3n-13)n/2∴n=1,Sn=5。n=2,sn=7n≥3,sn=14+(3n-13)n/2。
热心网友
解：f(x)=x^2-2(n+1)x+n^2+5n-7 = [x-(n+1)]^2 +(3n-8)顶点的坐标: (n+1,3n-8)an=n+1，所以{an}为等差数列∴bn=|3n-8|，设cn=3n-8，bn=|cn|当n≤2,cn=3n-8＜0,当n≥3,cn=3n-8＞0当n≤2,Sn =b1+……bn=-（c1+……cn）=-[-5+（3n-8）]n/2=（-3n^2+13n）/2当n≥3,Sn =b1+b2+b3+……+bn=-c1-c2+c3+c4+……+cn=c1+c2+……cn-2c1-2c2=[-5+（3n-8）]n/2-10-4=（3n^2-13n-28）/2
热心网友
f(x)=x^2-2(n+1)x+n^2+5n-7 = [x-(n+1)]^2 +(3n-8)顶点的坐标: (n+1,3n-8)(1). 显然，顶点的横坐标构成的数列{an}={n+1}为等差数列。(2). 顶点到x轴的距离构成的数列{bn}={|3n-8|}={5,2,1,4,7,....}n = 3时：Sn = (3*n^2 -13n+28)/2