ページ1:

1918 (1) k² = 3² + 4² + 5³ + √(6³
(2)
(3)
k=③
— j¸² = ³ ³ + 4² + 5 ³ + [6 ³
(+2) = (1+2)+(2+2)+ (3 + 2)² + (4+2)
3² + 4 ³ + 5² + 63
練 22
左辺
右辺 =
=
(2k+1)
(2+1)+(2·2+1) + (2-3 +1) + + (2-10 +1)
3 + 5+ 7 + + 21.
Σ(2-1)
i=2
=
(2-2-1)+(2-3-1) + (2-4-1) +-
+ (2-11-1)
=
3+5+7+ +21
左辺 = 右辺
E (2k +1) = — \2i-1).
k =
k=1
3=3+3+3+3=アル
hel
(2) 5 5+ 5+
+ 5 = 5n
h (B
Σ C = C+ C++ c = nc
k=1
NET
よって
Σ c = nc
1=1
Σ1 = n

ページ2:

P.92
-B 和の記号 (シグマ)
n
•Σ ak = a + a₂ + az + + an
k=0
Fak
9
初項(第1項)から第頃までの和」
22
例 7. (1)
さん=1+2+3+..+
(2)
k=00
E (3½-2) = (³ (1-2) + (3-2-2) + (3·3-2)+- + (3-2)
k =0
= 1 +4 + 7++ (34-2)
(3) 23= 30+3+38+ +
k-2
21 (1)(5k-1)
(2)
K-T
"
k-1
=
(5-1-1)+(5-2-1)+--
+ (52-1)
= 4+9+.. + (5n-1)
= 2+2
=
2+1
+1
+
+2
-
2"+1
ふ
6=3
(2k+1)=(2-3+1)+(2-4+1) +.
= 7 + 9 + + 31.
+(2-15+1)

ページ3:

P93
一数列の和の公式
4.
2.
Σ c = nc.
k = 1
k = n (n+1)
(++... + (=nc
18
1+2+ + n = 1/1/n (n+y).
3.
±² k² = 1 n
h (n + 1) (2n+1)
1+2+
++ n² = {n (n+1) (2n+1)
12
4.
k²-{+}
5. ray-
n
k = 1
練23 (1) 11/19 (1011)=511-
=
公式2.
= 55
11=10
a=1. r, na = cc 2231940
Sn = a(-1)-1-(1-1)
r-T
E k² = f · 2 · ( ε + 1 ) ( 2 + 1 ) = { 1-17
(2)
k=1
公むろ
n=8
=204.
(3)
公式5
h=6.
r = 2
=64-1=63,

ページ4:

THE HE NE
ak
a 9
Zak
たが1かられまでの
(a, as an 2(9)
和

ページ5:

1.93
この性質
(a₁ + b₁) + (ax+ bx) + + (an + hm) = (a + a2+-+ an) + (h₁th₂+ + bul
=
(anthelas + ba
pa, + paz++ pan
Pak
-=
ヱの性質
Pla₁ aut.. + A3)
I Σ (ax+b₂) = [ak + Elk.
2. Ipak
=
Σak
P
Σ (pau + & ba)
Σ pan + Σ q ba
(公式より)
Σ Ah + q [ b h
12525)
Σ (p. ac + q hk) = p Σ A a + q Σ bh
Σ (ae - ha) = Σan - [ bh

ページ6:

P.94
例 9.
=
=
=
=
-
=
2=1
(k-sk+2)
Ch² - 32k +42
M
11n(n+1)(2mtl)-3-12 n(ntl) +2n
1 (1匹の性質
5 n ( n + 1 ) ( 2 n + 1) - of 1) (n + 1) + 1 n
//{(n+1)(2n+1)- 9(n+1)+12}
= f f n ( 2 n ² + n + 2n+1 - 9n-9+12).
= f h ( 2 n ² - b n + 4 )
n (n² - 3 n + 2).
11/13n(n-1)(n-2)
[2] 和の公
[ [3] 通分
↓ [4] 共通
↓ [5] 展開
1161回分解
11
24(1)(+3)
↓[1]の性質
検
4Ck+C3
4.11n(n+1)+3m
= 2m (n+1) +3
{2(n+1)+3}
= n (2n+2+3)
=n (2n+5)
[2]和の公む
→
2-1982
初質
(4k+3)
4.1+3=7
(2n+5)-1-(2-1+5)

ページ7:

P.94
191.886
1-2 + 23 +34 +
+
13(n+1)
a= 12
(+1)
a = 2² 3-2 (2+1)
a3 = 3²· 4 ·3 (3+1)
a = n² (n+1)
ak = k² (k+1) \
k=1からまでの和
E k² (k+1) = [(k²+ h²).
L=0
= Σ k² + ck²
= { // n ( n + 1 ) + ( | 1 ( n + 1) (2n+1)
=
= | | π ² ( n + 1 ) ² + ( n ( n+1) (2n+1)
12-200
=
nn (n+1)(n+1) +
12
巳の性質
↓ [2] 和の公式
(n+1) (2n+1)
203757
=
虎nintl){3n(n+1)+2(2n+1}}[4] 共通固
1/2 n ( n + 1 ) ( 3 π ² + 3n + 4n+2).
J15]&W
=
12 h (n+1) (3n² + 7n+2)
1/2 h (n+1) (n + 2).(3n+1)
↓[6]回分解
=1^2=2.
11=1a62
12. 1. (141) · (142)-(3-141) = 12·1·2·3-4=2

ページ8:

+ 2n (3n-1)
P.95
2-1 (3-4-1)
25 (1) 2-2+4.5+・・・
=Σ2k (k-1)
k =
=
Σ (6k-2k)
= 6Ľk² - 2Ľk
= 6. fn (n+1) (2n+1)-2 — — -n (n+1)-
= n(n+() (2n+1)-n(n+1)
= 1 (n + 1) { (2n+1)-15
= 2n² (n+1).
(2) (+13)+(2+2³) +...+ (n+n³)
=
-
·C (k+k²)
Ck + C k²
== n(n+1)+{{n(){²
=
=
=
½n (n+1) + n² (n+1)²
— n (n + 1) + — · n⋅n. (n+\)(n+\).
n
4
(n+1)/2+1
( n (n+1) (n²+ n + 2)

ページ9:

(2)(3μ-7k+4)
-
=
~
+44
3. fn(n+1) (2n+1)-(+) + 4n
~
½ 11 (n+1) (2n+1)(n+1)+
自田{(n+1)(2mtl)-7(n+1)+ -8 }
½n (2n² + n + 2n +1-7-7+8)
½n (2n²-4n+2)
½n (n² - 2n+1)
1034942
[2] 和の公む
1 [3] 通分
↓ (4) 共通因数
[5] 展開
=
n (n-1)²
(3) k (k+ 2)
k-T
Ck² + C2k
n (1+1) (2n+1) + 2 = n(n+1)
=
⋅ b (n+1) (2n+1)+
42 (n+1)
=
(n+1)(2n+1)+6}.
n(n+1)(2n+7)
↓ [6] 国分
カート
(4)5k
=
5Ck
k-1
k = (+1)
1-4+11
=
5(-1) (1-1)+1
= ½ (n-1)-n.
n(n-1)