ページ1:

<搞事啊?
Do Why 要有這種東西?
(不須要它好像
數學家”又“想要偷懒!於是將"連加”的概念用”符號”表示。
①
2.已知H2+3+…+n=
n(n+1)
2
© 1² + 2 ² + 3 ² + ... + n² = n(n+o (2n+1)
③
+ + n² = [ n[^\+1) }²
2
Note:所以我們的連加要以這個為基礎!!
① 從1開始加E
連續項相加
重要!
只能背
除非你想一個一個
慢慢加!!
<推理開始囉]
3.推論:怎麼達成上述兩個要求
1.要從1開始→不如就當成首项(第”1”项)造成
2° 連續项相加→接下來就當成0.03...连续
舉例
1. (0+11+12+1+20
(法工)將須要的列出來,再把多餘的扣掉
=
n=20
(1+2+1+9+10+m+20)
n=9
-(1+2+…+9)
20 x 21 9×10
2
這個不须要)
「講得那麼簡單 222)
(法Ⅱ)用數列連加的精神
我很推薦
令a=10.02=1103=12→公差d=1
則一般项am= 10+(n-1)x1=9+n
⇒原=(9+1)+(9+2)+(9+3)+n+(9+11)
=
10x11
9+9+1+9= 9x11+ 10!!
+1+2+1+10
1

ページ2:

來進階點囉.
2
(法])
1²+ 3 ² + 5² +++ 19²
n=19
1² + 2² + 3 ² +4² + 1 + 18² + 19² - (2² + 4² + 1 + 18²)
= S₁₁ = [(2×1} +(2×2} ++ (2x 9Ƒ³] = S₁q − [2* 1ª + 2ª× 2*+m+2*xq³]
-9
= [4] - 4x 9×10×19
S19 - 2² ( 1² + 2 ² + − + 9² ] = (9×20×39
(Ⅱ)
6
6
A₁ = 1², a₂ = 3², a₂ = 5² →‡ààíû¥d=2
→ An= (1+ (n-Dx2)² = (2n−1)² = 41²-4n+1
-
$2₁ = A + A₂ + A3+on+ais
2n−1=19 → n=15
= (4× 1²-4×1+ \)+ (4xZ-4x2+1)+(4×3²-4×3+1)++(4×15=-4×15+1)
= 4x1² + 4x2² + 4x3² + \\\ + 4×15²
-4×1
+1
-
4×2-4×3
+|
-
-
4×15
+
+1
121
· 4x (1²+ 2² + 3² + \\ + 15²) - 4× (1+2+3+1+15) +1 × 15
=
= 4x
15816831
15x16
+15
2
[分開來看吧!」
2

ページ3:

3. 1²+ 4² + 7² + \\ +22²
(法工). 要用扣的有點難......用新招試試看!
(STD).
≥ Q₁ = 1ª₁ Q₁₂ = 4. A₂ = 7² → ˆ¥d=3 → An=(1+ (n-1)x3) = (3n-2}
'
as
=91²-124+4
& IR 3n-2= 22 → n=8
+2 = a₁ + A2 + A3++ Ag
8+9+4×8
+...+9x8² = 9x8x9x17 4x 8x9 +4x8
= 9x1² + 9×2² + 9׳ ³² + 1 + 9x8²
-4x1 - 4x2 -4x3 + m -4×8
+4+4
+4 + +4
(欸!不從1開始
4. 5² + 9³ + 13 ³² +
+ 21³
Ca
也可以欸~
你們還活著嗎? 888
⇒
> > As² α²² 9²² ² → Q#d-4 » Q-(5+ (n-0×4)³
.
& *
(4n+13
= 64n²+48n² + (2n+1
4n+1=21 >n=5
⇒☆ Q> 64x(i²+2²³+«‹+ 5³)=64x (5x612
+48×(1²+2²+1+5²)
+468x5×6×11
+ (2x (1+2+m+5)
+12x 5×6
+1x5
+5
NEW TO "
精神有學
3

ページ4:

5. 1x3+2x5+3x7+ + 10×21
14
⇒ a₁ =1, a₂ = 2, A3 = 3, ... → An=
b₁ = 3, b₂ = 5, b₂ = 7... → bu=3+ (n-1)x2 = 2n+
⇒ Anx bn= n(2n+1) = zn²+n
$ = A₁x b₁+ Axb₂ + \\ + Oxoxbro
2x² +2x2² + n + 2×10°
=
+ 1
+2 +"
10x11x21
10x11
= 2x
+
6
Z
+ 10
6. 3×21+5×18+7+15 +1:15*3
⇒ ½ a₁ = 3. A₂ = 5,...
b₁ = 21.62=181
11
→ An = 3+ (n-1)x2 = 2n+1
→ bn = 21+ (n-1) x (-3)= −3n+24
⇒ Anx bu= (2n+1)(3n+24) = -6 n² + 45n+24
且末项 2n+1=15
⇒=-6×1²-6x2²
-
+45x1 +45x2 +
113
n = 7
-
- 6x 17² = 6 x 7x 8 x 15 + 45x27x8 + 24×7
+45×7
x6
+24
+24
111
724
7.1+(1+2)+(1+2+3)+==+ (H+2+3+m+15)
⇒ & A₁ = 1 ₁ A₂ = 1+2 Q1 An = (+2+3+4+n= n(n+1)= 1/2
故原+++
2
++
Z
15² = 74(15x16x31) + 1x 15x-16
4