Question

设数列{an}的前n项和为Sn，对n属于N*，都有an=5Sn+2成立，1.求数列{an}的通项公式。

Answer 1

an=5Sn+2
an-1 = 5Sn-1 + 2
an - an-1= 5Sn - 5Sn-1 = 5an
- an-1 = 4an
q = an/an-1 = -1/4

a1=5a1+2
a1=-1/2
an = -1/2*(-1/4)^(n-1)

Answer 2

a_n=5S_n 2
a_(n-1)=5S_(n-1) 2
a_n-a_(n-1)=5(a_n)
4a_n=-a_(n-1)
a_1=5S_1 2
a_1=-1/2
a_n=-1/2*(-4)^(n-1)