第1个回答 2016-07-31
(1)An=Sn-Sn_1
=n^2-(n-1)^2
=2n-1
(2)Bn=2/(2n-1)(2n+1)
=1/(2n-1) -1/(2n+1)
所以
Tn=B1+B2+B3+...Bn=(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/(2n-1)-1/(2n+1))
=1-1/(2n+1)
. =2n/(2n+1)
=n^2-(n-1)^2
=2n-1
(2)Bn=2/(2n-1)(2n+1)
=1/(2n-1) -1/(2n+1)
所以
Tn=B1+B2+B3+...Bn=(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/(2n-1)-1/(2n+1))
=1-1/(2n+1)
. =2n/(2n+1)
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