Question

已知Sn/n=3n-2（1）求an（2）bn=3/(an*a(n+1)),Tn是bn的前n项和.求使得Tn

Answer 1

1)

a(n)=S(n)-S(n-1)=n(3n-2)-(n-1)(3n-5)=6n-5

2)

bn=3/[(6n-5)(6n+1)]=[1/(6n-5)-1/(6n+1)]/2

故Tn=[1-1/(6n+1)]/210-10/(6n+1)>10

故m最小值为10