Question

已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*，证明{an-1}为等比数列

Answer 1

∵Sn=n-5an-85
∴an=Sn-S(n-1)=n-5an-85-(n-1)+5a(n-1)+85
=1-5an+5a(n-1)
∴6an=5a(n-1)+1
6an-6=5a(n-1)+1-6
6(an-1)=5(a(n-1)-1)
∴an-1=5/6*(a(n-1)-1)=5/6*b(n-1)
∴｛an-1｝为公比是5/6的等比数列，O(∩_∩)O，希望对你有帮助

Answer 2

Sn=n-5an-85
S1=1-5a1-85
即a1=1-5a1-85
解得a1=-14
an=Sn-S(n-1)
=n-5an-85-[(n-1)-5a(n-1)-85]
=-5an+5a(n-1)+1
6an=5a(n-1)+1
an=5a(n-1)/6+1/6
an-1=(5/6)[a(n-1)-1]
(an-1)/[a(n-1)-1=5/6
所以{an-1}是以-15为首项5/6为公比的等比数列
an-1=(-15)*(5/6)^(n-1)

Answer 3

Sn﹣₁=(n-1)-5an﹣₁-85
an=Sn-Sn﹣₁=n-5an-85-[(n-1)-5an﹣₁-85]
6an=5an﹣₁+1
6an-6=5a(n-1)+1-6
6(an-1)=5(5an﹣₁－1)
an-1=5/6*(5an﹣₁－1)
古{an-1}为公比是5/6等比数列

Answer 4

Sn=n-5an-85
Sn-1=n-1-5a(n-1)-85
an=Sn-Sn-1=1-5an+5a(n-1)
则 6an=5a(n-1)+1
∴ 6an-6=5a(n-1)-5
即 (an-1)/[a(n-1)-1]=5/6
所以，数列{an-1}是以5/6为公比的等比数列

Answer 5

a(1)=s(1)=1-5a(1)-85,6a(1)=-84,a(1)=-14.
a(n+1)=s(n+1)-s(n)=(n+1)-5a(n+1)-85-[n-5a(n)-85]=1-5a(n+1)+5a(n),
6a(n+1)=5a(n)+1,6a(n+1)-6=5a(n)-5,
6[a(n+1)-1]=5[a(n)-1],
a(n+1)-1=(5/6)[a(n)-1],
{a(n)-1}是首项为a(1)-1=-14-1=-15,公比为(5/6)的等比数列.

Answer 6

推荐答案挺好的