Question

已知数列{An}的前n项和为Sn,且Sn=n-5An-85,n属于正整数,求sn的通项公式及Sn的最小值

过程要详细

Answer 1

Sn = n-5an-85 (1)
put n=1
a1=1-5a1-85
a1=-14
S(n-1)= (n-1) -5a(n-1) - 85 (2)
(1)-(2)
an= 1-5an+5a(n-1)
6(an -1)= 5(a(n-1) -1)
(an-1)/(a(n-1)-1) = 5/6
(an-1)/(a1-1) = (5/6)^(n-1)
an=1 -15. (5/6)^(n-1)
= 1- 18. (5/6)^n
Sn = a1+a2+..+an
= n - 15(1-(5/6)^n)/(1-5/6)
= n - 75(1-(5/6)^n)
min Sn = S1
= a1
= -14