在数列an中,a1=1/5,an+an+1=5的n+1次幂分之六,求数列前n项和sn

Question

在数列an中,a1=1/5,an+【an+1】=5的n+1次幂分之六,求数列前n项和sn

Answer 1

an+a(n+1)=6/5^(n+1)=(5+1)/5^(n+1)=1/5^n+1/5^(n+1)
a(n+1)-1/5^(n+1)=-(an-1/5^n)
a1-1/5^1=1/5-1/5=0
an-1/5^n=0
an=1/5^n
a1=1/5
数列{an}是以1/5为首项,1/5为公比的等比数列.
Sn=(1/5)[1-(1/5)^n]/(1-1/5)=(4/5)(1-1/5^n)