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