Question

数列{an}的前n项的和Sn=n2-10n(n属于N*),数列{bn}满足bn=(an+1)/an(n属于N*),(1)

Answer 1

(1)

Sn =n^2-10n

an = Sn -S(n-1)

= (2n-1) -10

= 2n-11

=>{an}是等差娄列

(2)

bn = (an+1)/an

= (2n-10)/(2n-11)

max bn = b1 = 8/9

min bn = b5 =0

(3)

an > 0

2n-11 >0

n > 11/2

n= 6

cn =|an|

for n =6

Tn = -(a1+a2+..+a5) +(a6+a7+..+an)

=25 + (2n-11+1)(n-5)/2

=25 + (n-5)^2

=n^2-10n +50