Question

高中数学数列问题。，

Bn=3/An*A（n+1）
An=6n-5
求Bn的前N项和Tn

Answer 1

Bn=3/[(6n-5)(6n+1)]=1/2[1/(6n-5)-1/(6n+1)]
2b1=1/1-1/7
2b2=1/7-1/13
2b3=1/13-1/19
.
.
2bn=1/(6n-5)-1/(6n+1)
两边相加得:2Tn=1-1/(6n+1)
Tn=(3n-1)/1/(6n+1)

Answer 2

Bn=3/(6n-5)(6n+1)=1/2*[1/(6n-5)-1/(6n+1)
Tn=1/2*{1-1/7+1/7-1/13+~+1/(6n-5)-1/(6n+1)}
=1/2[1-1/(6n+1)]
=3n/(6n+1)

Answer 3

1/BN=AN*A(N+1)/3=(6N-5)*(6N+6-5)=(6N-5)*(6N+1)
BN=3/(6N-5)*(6N+1) =3*6*(1/(6N-5)-1/(6N+1))=18*(1/(6N-5)-1/(6N+1))