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

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

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======以下答案可供参考======供参考答案1: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供参考答案2:bn=3/[(6n-5)^2+1]tn根据欧拉公式∑1/n^2=π^2/6tntn=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

谢谢了