第二集

 1-2 等差級數

Series (級數) - According to wiki definition , 一個有窮或無窮的序列  的元素的形式和稱為級數.  所以如果將任意的數列 a0,a1,a2,a3,... , an , ... 全部的項目相加起來 即 a0+a1+a2+a3+... +an+ ... 稱之為級數. 
如果存在一個等差數列 a0,a1,a2,a3,... ,an 則將其所有的項目加起來所形成的一種形式 a0+a1+a2+a3+... +an , 稱之為等差級數 or 算術級數. 
因為等差數列 a0,a1,a2,a3,... ,an 其公差為 d 則
a1+a2+a3+... +an = a1 + ( a1 + d ) + ( a1 + 2d ) + ( a1 + 3d ) + ... + ( a1 + nd ) = (n+1) a1+ (1+2+3+...+n)d  
此時令 Sn = a1+a2+a3+... +an , Sn 為其等差級數之和 且 Sn  = [(a1+an)n]/2 
證: 
取任意一個等差級數  a1+a2+a3+... +an  且令  Sn = a1+a2+a3+... +an 
因為 a2 = a1 + d , a3 = a1 + 2d , a4 = a1 + 3d , ... , an = a1 + (n-1)d 則 
Sn =  a1  +  (a1+d)  +  (a1+2d)  +... + (a1+(n-1)d) 我們可以將其反過來加總,
Sn =  (a1+(n-1)d) + ...                     +  (a1+d) + a1 

將以上兩式相加,可以得到如下的式子: 
2Sn = 2a1+(n-1)d + 2a1+(n-1)d  + ... + 2a1+(n-1)d 
2Sn = (a1 + an) + ... + (a1 + an) , 共有 n 項 ( a1 + an) 
2Sn = n(a1 + an) 
Sn = n(a1 + an)/2  = n/2 [ a1 + a1 +(n-1)d ] = n[ 2a1 + (n-1)d ] / 2    Q.E.D. 

考量部分和如下: 
S1 = a1 
S2 = a1  + a2
S3 = a1  + a2 + a3 
.... 
Sn-1= a1 + a2 + a3 + ... + an-1 
Sn =  a1 + a2 + a3 + ... + an-1 + an 
觀察如上的結果,可以表示如下的方式: 
S1 = a1 
S2 = S1 + a2
S3 = S2 + a3 
.... 
Sn = Sn-1 + an , that is Sn is a recursive definition , when n ≧ 2 , n is integer . why?  Because Sn is a recursive function , then it must owns a recursive base . If n =1 , then s0  is undefined .  

Basically, a series is a function. Obviously , we regard it as n functions addition.
Let f(n) = an , n =1,2,3, ... then a series that has n terms from  1 to n . 
Sn =  a1 + a2 + a3 + ... + an-1 + an  = f(1) + f(2) + f(3) + ... + f(n)
f is a function and A is its domain and co-domain B , f : A → B and h is also a function and B is its domain and C is its co-domain, then g : A → C , g = h ∘ f 
According to above property , Sn is a function from domain A to co-domain C . A is Nature numbers set N . 
Some series properties 
1. 等差級數的和 = 項數 * 中央項  即 Sn = n(a1 + an)/2
2.第n項為前n項的和-前n-1項的和 ,即 an =  Sn -  Sn-1 
Consider Sm and Sn , m < n 
Sm =  a1 + a2 + a3 + ... + am 
Sn  =  a1 + a2 + a3 + ... + am + am+1 + ... + an 
Sn - Sm = am+1 + ... + an 
Sm-1 =  a1 + a2 + a3 + ... + am-1 
Sn - Sm-1 = am + am+1 + ... + an 

Ex. 存在一個等差級數 a1 + a2 + a3 + ... + a8 , 其 difference d 是3 且第四項是7 , 求 S8 
S8  = a1 + a2 + a3 + a4... + a8 = 8/2(a1+a8) = 4 ( a1 + a1 + 7*3) = 84 + 8 a1 
a4 = 7 =  a1 + 3d = a1 + 9 , a1 = -2 ∴ S8 = 84 - 8*2 = 68  
Ex. Assume there exists an arithmetic series a1 + a2 + a3 + ... + an and 1st term is 5 and last term or nth term is -28 and its difference d is -3 , find how many terms and Sn 
We assume an arithmetic series  a1 + a2 + a3 + ... + an and its summation 
Sn = ∑ ai , i =1,2,3, ... 
 an = a1 + (n-1)d  
-28 = 5 + (n-1)(-3) 
-28 = 5 -3(n-1) 
-28 = 5 - 3n + 3 
-28 = 8 - 3n 
-36 = -3n 
n = 12 , So S12 = 6*(5-28) = -138 


Let's see another example as below: 
There exists a function f(n) = 200 -3n , n is a nature number . Find the followings 
1. The value of n that causes the summation of f(1)+f(2)+f(3)+ ... + f(n) is 
    maximum
2.  the maximum value of f(1)+f(2)+f(3)+ ... + f(n) 

Given an arithmetic series a1 + a2 + a3 + ... + an and ai = f(i) , i =1,2,3,... 
and f(n) = 200-3n . 
a1 = f(1) = 200 - 3 = 197 
a2 = f(2) = 200 - 6 = 194 
a3 = f(3) = 200 - 9 = 191 
... 
an = f(n) = 200 - 3n = 200-3n  

題意是如何找到n 值使得 f(1)+f(2)+ ...+ f(n) 其和為最大 ? 
We can discover a fact , that is f(1) > f(2) > f(3) > .... 所以每一項是慢慢地遞減的. 但是要到多少項之後所有的項都是負的 ? 發現其公差為 -3  ,所以必然有一項之後會開始為負的.
假設那項是第k項之後為負 , ak > 0 
200 -3k > 0 
200 > 3k 
200/3 > k 
k < 67 , 即 k= 67 或稱從第 67 項開始之後每項是負的.  所以考慮部分和
S1 = a1 = f(1) > 0 
S2 = a1 + a2 = f(1) + f(2) > S1
S3 = a1 + a2 + a3 = f(1) + f(2) + f(3) > S2
S4 = a1 + a2 + a3 + a4 = f(1) + f(2) + f(3) + f(4) > S3 
... 
S66 = f(1) + f(2) + f(3) + ... + f(66) >  S65  > S64 > ... So n = 66 causes Sn is maximum. 
Again,  see another example as below . 
ex. 有兩個數字 2 與 59 , 在其中插入n個數 ,使其成為等差數列,而n+2個數的總和為610 求插入數中的最大數為何 ? 
Assume there exists an arithmetic series a1 + a2 + a3 + ... + an+1 + an+2 , 
a1 = 2 and an+2 = 59 
Sn+2 = (n+2)(2+59)/2 = 610 
   1220 = (n+2)(61) 
   20 = n+2 
 n = 18 , So this series has 200 terms , they are  a1 + a2 + a3 + ... + a20  
Again , we start calculating the difference d . 
a20 = a1 + (20-1)d 
59 = 2 +19d 
57 = 19d , d = 3 
a1 = 2 
a2 = 5 > a1
a3 = 8 > a2
... 
a20 = 59  > a19  ∴ a20 > a19 > ... > a1 , So the maximum term of inserting numbers is a19 (56). 

1 - 3 等比數列 
What is it ?

Geometric sequence (G.P.) -  a geometric sequence , is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio . 
Assume that there exists a sequence , a1,a2,a3, ...,an , and  ai+1 / ai = r , r ≠ 0  
,called it is a geometric sequence and if add them all , likes a1+a2+a3+ ...+an , then called it is geometric series .   
a2 = a1 * r  
a3 = a2 * r  
a4 = a3 * r   
... 
an = an-1 * r  = ( an-2 * r ) * r = (( an-3 * r ) *r ) *r ... = a1 * r ^n-1  
如果我們將每個項視為一個量,則每個項與相鄰項都成一個比率(稱之為公比) r,由幾何角度就是圖形的放大及縮小, 故稱之為幾何數列 . Please see figure below . 

Fig - 幾何數列圖形

Consider above figure , assume the side of square is 1 unit , then its area is 1. 

If we sum up those partitions , the summation of them 1/4 + 1/16 + 1/64 + ... + 1/2 ²ⁿ + ... It is a Geometry series and its ratio r = 1/4 .
Let its summation of first n terms is Sn , then Sn = 1/4 × ( 1 + 1/4 + 1/16 + ... + 1/2 ^2n-2 )  4Sn  =  1 + 1/4 + 1/16 + ... + 1/2 ^2n-2  ; a1 = 1  ,  a2 = 1/4 , a3 = 1/16 , ... , an = 1/2 ^2n-2  . r = 1/4 , then 4Sn = 1/(1 - 1/4) = 1/3/4 = 4/3. What's wrong ? 
(1/4 + 2 × 1/4 ) + ( 1/16 + 2 × 1/16 ) + ... + (  1/2 ^2n-2 + 1/2 ^2n-1  ) = 3 × ( 1/4 + 1/16 + ... +  1/2 ^2n-2    )  = 3/4 × ( 1 + 1/4 + ... +  1/2 ²ⁿ ) , Why ? Same as above. 
我們可以歸納一個現象: 
即n代表分切正方形的次數, 
n = 1 , 4 × 1/4
n = 2 , 3 × 1/4  + 4 × 1/16 = 3/4 + 1/4 = 1
n = 3 , 3 × 1/4  + 3 × 1/16 +  4 × 1/64  = 3/4 + 3/16 + 1/16 = 1
... 以此類推
若 n = k , k >> 1 ; 則 令 p(n) 為一個命題計算,即一個正方形邊長分割k次的面積總和為Sn , Sn =  3 × 1/4  + 3 × 1/16 +  3 × 1/64 + ... + 4 × 1/4 ⁿ + ...
Sn = 3 ×  ( 1/4 + 1/16 + 1/64 + ... + 1/4 ⁿ ) + 1/4 ⁿ  

令 Ai =  (1/4) ⁱ     所以 , Sn = 3 ×  (A1 + A2 + A2 ... + An  ) + 1/4 ^n 
所以,    Sn = 3 ×(A1 / 1-1/4 )= 3 ×(1/4  × 4/3) = 3 × 1/3 = 1 , because when n is very large or infinite large , 1/4 ⁿ  approaches zero .  

More  information about Geometry series , please refer to Geo Series and Geometry series