设bn=3/(anan+1),an=6n-5,tn是数列｛bn}的前n项和,求使得Tn

bn=3/[(6n-5)^2+1]
tn根据欧拉公式∑1/n^2=π^2/6
tntn=1.5+3/50+∑3/[(6n+7)^2+1]>1.56+∑3/(2n+3)^2
上式右端=1.56+1/3[-1/9-1+∑1/(2n-1)^2]
=1.56+1/3[-1/9-1+∑1/n^2-∑1/(2n)^2]
=1.56+1/3[-1/9-1+3/4∑1/n^2]
=1.56+1/3[-1/9-1+3/4*π^2/6]
=1.60086>32/20
故m=33

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) 所以m最小取10

Tn
=b1+b2+...+bn
=(3/a1a2)+.+3/[ana(n+1)]
=3[1/a1a2+1/a2a3+...+1/ana(n+1)]
=3[1/(1*7)+1/(7*13)+...+1/(6n-5)(6n+1)]
=3{(1/6)(1-1/7)+(1/6)(1/7-1/13)+...+(1/6)[(1/6n-5)-1/(6n+1)]}
=(1/2)*[1-1/7+1/7-1/13+.+1/(6n-5)+1/(6n+1)]
=(1/2)*[1-1/(6n+1)]
因为n属于N*
所以1/(6n+1)>0
则:
Tn=(1/2)-(1/2)[1/(6n+1)]=10
所以
最小正整数m为10