n
(3)
(3)
(5)
ay=10
a-5=
n
aq = 1
a
n
Aan-
n-1
+5,n≥2
E: (2) (3) (5) 14h4') = Dar
0
(4)
ap
a
n
100
-5=(a-1-5), n≥2
An-2=-41am-1-2) spivs
An-5=-419-1-574
V
2
解:(1)an
[2
2+2+2~
+23-n
2"-1
=1-
收斂於4
2
=
4
立
【雄中】
(2)
(3)
(4)
(5)
a = 5
((a,-5)=-4(a-1-5), n≥2'
aq=10
-5-(-5), 22
5=
a
n
1
a =
an
a =1
100
n-
S=(a+1-5), n≥2
#a
715-410-7
Axx=-4 (urs)
a =(-4)"-1 (a,-5)+5.5
n-1
an-5=-414+ 3
-(+) ** (₁, -5)+5 Jann's
n-1
s) 收斂於5
ans:
= ( 3 ) " (a₁ - 5 ) + 5 ·• h
故an
,
<a>為遞增數列,且有上界
n
a
n
4an-1
+5,n≥2
設收歛於a→a=v4a+5⇒a -4a-5=0-
azl
→ a = 5
做
15