高中数学数列

显示全部楼层 · 2009-6-6 15:44:48

因为点（n,Sn)在函数Y=3X^2-2X的图像上所以Sn=3n^2-2n这是一个等差数列的和因为等差数列前n项和Sn=(d*n^2)/2+(a1-d/2)n所以d/2=3,d=6,a1=1 an=1+6(n-1)=6n-52.bn=3/（an×an+1)Tn=3(1/a1*a2+1/a2*a3+.......+1/an*a(n+1))=3[1/a1(a1+6)+1/a2(a2+6)+.......+1/an(an+6)]=3{[1/a1-1/(a1+6)]/6+[1/a2-1/(a2+6)]/6+......+[1/an-1/(an+6)]/6}=1/2*{[1/a1-1/(a1+6)]+[1/a2-1/(a2+6)]+......+[1/an-1/(an+6)]}=1/2*[1/a1-1/(a1+6)+1/a2-1/(a2+6)+......+1/an-1/(an+6)]=1/2*[1/1-1/7+1/7-1/13+......+1/an-1/(an+6)]=1/2*(1-1/(an+6)=1/2*(1-1/6n+1) =3n/(6n+1)=3/(6+1/n)<1/2<m/20所以m最小取10

千问 · 2009-6-6 15:44:48

(1)∵点(n,Sn)在函数y=3x^2-2x上∴Sn=3n^2-2n,Sn-1=3(n-1)^2-2(n-1)∴an=Sn-Sn-1(n≥2)=6n-5当n=1时，Sn=1，满足上式综上得,an=6n-5(2)bn=3/an×an＋1=3/(6n-5)(6n＋1)=1/2[1/(6n-1)-1/(6n＋1)],得Tn=1/2[1-1/(6n-5)]又∵lim(1/6n-5)=0,∴limTn=1/2,所以Tn＜1/2,∴m的最小值为10