设bn=3/(anan+1),an=6n-5,tn是数列{bn}的前n项和,求使得Tn
人气:261 ℃ 时间:2019-10-31 18:12:17
解答
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
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