设bn=3/(anan+1),an=2n-51,tn是数列{bn}的前n项和,求使得Tn
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设bn=3/(anan+1),an=2n-51,tn是数列{bn}的前n项和,求使得Tn
![设bn=3/(anan+1),an=2n-51,tn是数列{bn}的前n项和,求使得Tn](/uploads/image/z/14935282-34-2.jpg?t=%E8%AE%BEbn%3D3%2F%28anan%2B1%29%2Can%3D2n-51%2Ctn%E6%98%AF%E6%95%B0%E5%88%97%EF%BD%9Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82%E4%BD%BF%E5%BE%97Tn)
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
=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
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