ページ3:

(2)
(1)より an
=3n-2
=
→ak 3k-2, ak+2 = 3(k +2)-2=3k+4
†³½³ ααk+2 = (3k − 2)(3k + 4) = 9k² + 6k − 8
n
-
n
よってΣ0.9k+2=Σ(9k2 + 6k-8)
k=1
k=1
1
であるから lim
n→∞
n
3
=
=
+6x/m
: 9×−n(n+1)(2n+1)+6×−n(n + 1) −8n
3
2
6
(2n³ + 3n² + n) + 3(n² + n) −8n
15
=
=3n³ +
n²
2
n
k=1
a = lim
akak
k+2
7
2
n
1
15
(3n³ +
2
n
3
n→∞
n
2
=
n→8
lim (3+
15
7
2n
2n²
7-2
n)
=3+0-0
=3圈

ページ4:

a=3n-2 より
a = 3k-2
ak+1
よって
(ak+2)² - (ak)²
2
ak+1=3(k+1)− 2 = 3k+1
ak+2=3(k+2)-2=3k+4
=
=
=
ak+1
(ak+2 +ak) (ak+2 - αk)
3k+1
(3k+4+3k-2)(3k+4−3k+2)
3k+1
(6k+2)×6
3k+1
(3k+1)×12
1
答
12
よりak+1=
ak+1
(a+2)-(a)
n
よって lim
n→∞
12
ak+1
(a)² (a+2)
k=1 k
(ax+2)² - (a)²
12
lim
n
ak+1
(a)²(a)²
n→∞
k=1
k
=
lim
n
12
k+2
(a+2)² - (ak)²
(ak)² (ak+2)²
部分分数分解
=
lim
1
=
1
=
n→∞
k=1
lim 1
n→ 12
k=1
((ak) ² (a + 2)²
1
1
H G D G D C D
n→ 12
a2 = 3×2-2
lim
1
n→ 12
2
a
+
2
a
- G
a
n-I\\
2
1
1
an+1
2
+
G
1
a
(3n+1)² (3n+4)² |
1
An+2
2
=
+
12
42
1+
0
12
16
17
192