设二次函数f(x)=x²+2x,x∈[n,n+1](n∈N*),f(x)的最大值与最小值之差为g(n).①求g

1个回答

  • 1:因为n∈N*,因此n>0.又因为f(x)=x²+2x,是一个过(0,0)和(-2,0)点,开口向上的抛物线,因此当x>0时,f(x)单调递增.

    所以,g(n)=f(n+1) - f(n)= (n+1)^2 + 2*(n+1) - n^2 -2*n = 2*n +3

    2:因为bn = (2*n+3)/2^n

    所以Tn = 5*(1/2) + 7*(1/2)^2+ ...+(2n+3)*(1/2)^n

    Tn/2 = 5*(1/2)^2 + 7*(1/2)^3+ ...+(2n+3)*(1/2)^(n+1)

    做差:Tn - Tn/2 = 5*(1/2)+2*(1/2)^2+...2*(1/2)^n-(2n+3)*(1/2)^(n+1)

    有:Tn/2 = 2.5 +[0.5 +0.5^2 +...+0.5^(n-1)] -(2n+3)*(1/2)^(n+1)

    = 2.5+0.5(1-0.5^(n-1))/(1-0.5) -[(2n+3)*(1/2)^(n+1)]

    = 2.5+1-0.5^(n-1)-[n*0.5^n+3*0.5^(n+1)]

    = 3.5-(n+3.5)0.5^n

    =7/2 -(n+7/2)*2^(-n)

    3:因为an=(2n^3-3n^2)/(2n-3)= n^2

    且a(n+1)-an = 2n+3

    所以:Sn=a1-a2+a3-a4+.+(-1)^(n-1) a(n-1)

    =1-4+9-16.+(-1)^(n-1)a(n-1)

    两两配对,可以得到:

    Sn = a1+(-a2+a3)+...+()

    =a1+(3^2-2^2))+.

    讨论:

    当n为奇数时:

    Sn = a1+(3^2-2^2))+.+(-(n-1)^2+n^2))

    = 1+5+9+...(2n-1)--------------------》这是个公差为4的等差数列

    =n(n+1)/2

    当n为偶数时:

    Sn = (1^2-2^2)+(3^2-4^2)+...+[(n-1)^2 - n^2]

    = -(3+7+11+...+(2n-1))--------------------》这是个公差为4的等差数列

    =-n(n+1)/2