型二 : 等比中項
等比中項(幾何中項) 如何求 ?
假設有一等比數列 a,b,c , 及其公比是 r , r ≠ 0 則由性質, 可以得 b/a = c/b
ac = b2 , b = ±√ac
EX. There exists a G.P and 4th term is 6 , i.e. a4 = 6 and a1 + a2 = 12 × a3 , find the G.P.
Assume there exists a G.P. , a1 , a2 , a3 , ... , an , and its common ratio r , r ≠ 0 .
According to the property , an = a1 × r n-1 , then a4 = 6 = a1 × r 3
a1 + a2 = a1 + a1 × r = a1 × (1+r) = 12 × a1 × r 2
1+r = 12 × r 2
1+r -12r 2 = 0 , (4r+1)(-3r+1) = 0
⇒ r = -1/4 or 1/3 then
等比中項(幾何中項) 如何求 ?
假設有一等比數列 a,b,c , 及其公比是 r , r ≠ 0 則由性質, 可以得 b/a = c/b
ac = b2 , b = ±√ac
EX. There exists a G.P and 4th term is 6 , i.e. a4 = 6 and a1 + a2 = 12 × a3 , find the G.P.
Assume there exists a G.P. , a1 , a2 , a3 , ... , an , and its common ratio r , r ≠ 0 .
According to the property , an = a1 × r n-1 , then a4 = 6 = a1 × r 3
a1 + a2 = a1 + a1 × r = a1 × (1+r) = 12 × a1 × r 2
1+r = 12 × r 2
1+r -12r 2 = 0 , (4r+1)(-3r+1) = 0
⇒ r = -1/4 or 1/3 then
if r = 1/3 , then 6 = a1 × (1/3) 3
a1 = 27×6 = 162 , if r = -1/4 , then 6 = a1 × (-1/4) 3 ⇒ a1 = - 384
等比及等差數列 , 1, -1,1,-1,1,-1, ... 此為交錯數列, 如果將其所有項目加總則, 1-1+1-1+1-1+1.... 請問其和為多少 ? Basically , 若項數為奇數,即 n = 2m + 1 , m = 1,2,3,4, .... 則和為 1 , 若項數為偶數擇期和為 0 .
Some properties :
1. 等比級數對加減不滿足封閉性
For example ,
若有一個等比級數 1+2+4+8+16+32, ... 則我們對其所有的項皆加/減一個常數k ,則使得原級數 為一個新的級數 (1+k)+(2+k)+(4+k)+(8+k)+(16+k)+... ,如果 k ≠ 0 , 則 ai+1 / ai ≠ 2 , for example , if k =1 , then above G.P. can be expressed as 2+3+5+9+17+... 3/2 ≠ 5/3 ≠ 9/5 ≠ 17/ 9 ...
2. 等比級數對乘除滿足封閉性
Pf:
Assume that there exists a G.P. a1+a2 + a3 + ... + an , and its common ratio r . multiply a constant k , such that above G.P. as ka1 + ka2 + ka3 + ... + kan , then it still is a GP and common ratio still is same.
3. a,b,c,d 為等比數列, 則 ad = bc
pf:
假設 a,b,c,d 的公比為 r , 則 b/a=c/b=d/c=r → b/a = d/c ∴ ad = bc
4.a,b,c,d,e 為等比數列 , 則 ae = bd = c 2
pf:
假設 a,b,c,d,e 的公比為 r , b/a = c/b = d/c = e/d = r → c/b = d/c 且 b/a = e/d
∴ ae = bd = c 2
Assume that there exists a G.P. , a1,a2,a3, ...,an , then a1an = a2an-1 = ... = a⌈n/2⌉ × a⌈n/2⌉ , ∀n and n is odd .
EX. a,b,c 為等比數列, 且abc = 216 , 則 b = ? 若 ac = 81 求 b
∵ a,b,c 為等比數列 , 則 b/a = c/b = r ; r 是公比 , r ≠ 0
已知 abc = 216 則 ac = 216/b
216 = b 3 ∴ b = 6 同理 if ac = 81 , 則 b = 9 and -9 .
See another example as below :
There are two numbers 3 and 3072 . Now we insert n terms between them and causes it becomes a G.P. with n+2 terms . if the 4rd term is 192 , then what is the value of n ?
To resolve this question , we can assume that there is a G.P. as following
3 , a1 , a2 , a3 , ... , an , 3072 , and a1 , a2 , a3 , ... , an are those terms we inserted . Then we rewrite above G.P. again as below
b1 , b2 , b3 , ... , bn , bn+1 ,, bn+2 ; b1 = 3 , bn+2 = 3072 , bi+1 = ai , i ∈ { 1,2,3,4,...,n} and its common ratio is r .
等比及等差數列 , 1, -1,1,-1,1,-1, ... 此為交錯數列, 如果將其所有項目加總則, 1-1+1-1+1-1+1.... 請問其和為多少 ? Basically , 若項數為奇數,即 n = 2m + 1 , m = 1,2,3,4, .... 則和為 1 , 若項數為偶數擇期和為 0 .
Some properties :
1. 等比級數對加減不滿足封閉性
For example ,
若有一個等比級數 1+2+4+8+16+32, ... 則我們對其所有的項皆加/減一個常數k ,則使得原級數 為一個新的級數 (1+k)+(2+k)+(4+k)+(8+k)+(16+k)+... ,如果 k ≠ 0 , 則 ai+1 / ai ≠ 2 , for example , if k =1 , then above G.P. can be expressed as 2+3+5+9+17+... 3/2 ≠ 5/3 ≠ 9/5 ≠ 17/ 9 ...
2. 等比級數對乘除滿足封閉性
Pf:
Assume that there exists a G.P. a1+a2 + a3 + ... + an , and its common ratio r . multiply a constant k , such that above G.P. as ka1 + ka2 + ka3 + ... + kan , then it still is a GP and common ratio still is same.
3. a,b,c,d 為等比數列, 則 ad = bc
pf:
假設 a,b,c,d 的公比為 r , 則 b/a=c/b=d/c=r → b/a = d/c ∴ ad = bc
4.a,b,c,d,e 為等比數列 , 則 ae = bd = c 2
pf:
假設 a,b,c,d,e 的公比為 r , b/a = c/b = d/c = e/d = r → c/b = d/c 且 b/a = e/d
∴ ae = bd = c 2
Assume that there exists a G.P. , a1,a2,a3, ...,an , then a1an = a2an-1 = ... = a⌈n/2⌉ × a⌈n/2⌉ , ∀n and n is odd .
EX. a,b,c 為等比數列, 且abc = 216 , 則 b = ? 若 ac = 81 求 b
∵ a,b,c 為等比數列 , 則 b/a = c/b = r ; r 是公比 , r ≠ 0
已知 abc = 216 則 ac = 216/b
216 = b 3 ∴ b = 6 同理 if ac = 81 , 則 b = 9 and -9 .
See another example as below :
There are two numbers 3 and 3072 . Now we insert n terms between them and causes it becomes a G.P. with n+2 terms . if the 4rd term is 192 , then what is the value of n ?
To resolve this question , we can assume that there is a G.P. as following
3 , a1 , a2 , a3 , ... , an , 3072 , and a1 , a2 , a3 , ... , an are those terms we inserted . Then we rewrite above G.P. again as below
b1 , b2 , b3 , ... , bn , bn+1 ,, bn+2 ; b1 = 3 , bn+2 = 3072 , bi+1 = ai , i ∈ { 1,2,3,4,...,n} and its common ratio is r .
Because 4th term is 192 , i.e. b4 = 192 and 192 = 3 × r 3 , then r = 4 .
Now we have three known , they are last term 3072 , first term 3 and common ration r is 4 .
3072 = 3 × 4 n+1
2 10 = 2 2n+2
n = 4
Ex. 有兩數 a ,b 其等差中項為 10 , 等比中項為6 ,且a > b , 求 2a+3b
20 = a+b , 36 = ab .
36 + 20x + x2
(x - 18)(x - 2)
x = 2 or x = 18
∵ a > b
∴ b = 2 , a = 18
存在一個等比級數, a1+a2+a3+...+an , 其Sn = ?
∵ a1+a2+a3+...+an 是等比級數 , 令其公比是 r , 所以
a2 = a1 × r
a3 = a2 × r = a1 × r 2
a4 = a3 × r = a1 × r 3
...
an = an-1 × r = a1 × r n-1
Now we have three known , they are last term 3072 , first term 3 and common ration r is 4 .
3072 = 3 × 4 n+1
2 10 = 2 2n+2
n = 4
Ex. 有兩數 a ,b 其等差中項為 10 , 等比中項為6 ,且a > b , 求 2a+3b
20 = a+b , 36 = ab .
36 + 20x + x2
(x - 18)(x - 2)
x = 2 or x = 18
∵ a > b
∴ b = 2 , a = 18
存在一個等比級數, a1+a2+a3+...+an , 其Sn = ?
∵ a1+a2+a3+...+an 是等比級數 , 令其公比是 r , 所以
a2 = a1 × r
a3 = a2 × r = a1 × r 2
a4 = a3 × r = a1 × r 3
...
an = an-1 × r = a1 × r n-1
Sn = a1+a2+a3+...+an = a1 + a1 × r + a1 × r 2 + a1 × r 3+ ...+ a1 × r n-1
= a1 ( 1 + r + r 2 + r 3+ ... + r n-1 )
Consider that a function f(x) , f(x) = (1-xn ) × (1-x)-1 then above expression can be wrote as following :
Sn = a1 × (1-rn) × (1-r)-1
And another way to calculate Sn as followings .
Sn = a1 + a1 × r + a1 × r 2 + a1 × r 3+ ...+ a1 × r n-1
rSn = a1 × r + a1 × r 2 + a1 × r 3 + a1 × r 4+ ...+ a1 × r n-1 + a1 × r n
(r-1)Sn = a1 × r n - a1 = a1 × ( r n -1)
∴ Sn = a1 × (1-rn) × (1-r)-1 = a1 × [ ( r n -1) / (1-r) ] , | r | ≠ 1 , | r | < 1
if r =1 , above summation of n terms , Sn = a1 + a1 + ... + a1 + ... + a1 = na1
Ex. 一個等比級數 5/4+1/2+1/5+ ... , 求其第5項之和 .
由題意得知為一個等比級數,故可以令 a1 = 5/4 , a2 = 1/2 , a3 = 1/5 , ...
求其公比 r = ai+1 / ai , i = 1,2,3,4, ... ; therefore , r = 1/2/5/4 = 1/2 × 4/5 = 2/5. 及套用公式 Sn = 5/4 ×[1- (2/5)n ] / (1-2/5) , 所以S5 = 5/4 ×[1- (2/5)5 ] / (1-2/5) = 1031/500
Ex. 一個等比級數共有6項, a1 = 3 , a3= 375 , 求其公比及其之和 .
已知第一項 a1 = 3及第三項 a3 = 75 , 假設公比為 r , | r | ≠ 1.故
a2 = 3r
a3 = a2×r = 3×r 2 = 75 , r 2 = 25 , r = ± 5 .
所以此級數可以寫成 3 + 15 + 75 + 375 + 1875 + 9375 , when r = 5
或 3 - 15 +75 - 375 + 1875 - 9375 , when r = - 5 .
則 S6 = 3 × ( 5 6 - 1 ) / (5-1) = 3/4 ×(5 6 - 1) or S6 = 3 × ( (-5) 6 - 1 ) / (-5-1) = -3/6 ×(5 6 - 1) = (1- 5 6)/2
一些例子
0,0,0,0,....,0,0,0,0,... 為等差數列, 但非等比級數(因為公比為0/0 , 同時各項為0 )
1,1,1,1, ... , 1,1,... 為等差亦為等比數列
1,-1,1,-1,1,-1,..., (-1)n ,... 為正負交錯的交錯級數,也為等比級數
Sn = a1 × (1-rn) × (1-r)-1
And another way to calculate Sn as followings .
Sn = a1 + a1 × r + a1 × r 2 + a1 × r 3+ ...+ a1 × r n-1
rSn = a1 × r + a1 × r 2 + a1 × r 3 + a1 × r 4+ ...+ a1 × r n-1 + a1 × r n
(r-1)Sn = a1 × r n - a1 = a1 × ( r n -1)
∴ Sn = a1 × (1-rn) × (1-r)-1 = a1 × [ ( r n -1) / (1-r) ] , | r | ≠ 1 , | r | < 1
if r =1 , above summation of n terms , Sn = a1 + a1 + ... + a1 + ... + a1 = na1
Ex. 一個等比級數 5/4+1/2+1/5+ ... , 求其第5項之和 .
由題意得知為一個等比級數,故可以令 a1 = 5/4 , a2 = 1/2 , a3 = 1/5 , ...
求其公比 r = ai+1 / ai , i = 1,2,3,4, ... ; therefore , r = 1/2/5/4 = 1/2 × 4/5 = 2/5. 及套用公式 Sn = 5/4 ×[1- (2/5)n ] / (1-2/5) , 所以S5 = 5/4 ×[1- (2/5)5 ] / (1-2/5) = 1031/500
Ex. 一個等比級數共有6項, a1 = 3 , a3= 375 , 求其公比及其之和 .
已知第一項 a1 = 3及第三項 a3 = 75 , 假設公比為 r , | r | ≠ 1.故
a2 = 3r
a3 = a2×r = 3×r 2 = 75 , r 2 = 25 , r = ± 5 .
所以此級數可以寫成 3 + 15 + 75 + 375 + 1875 + 9375 , when r = 5
或 3 - 15 +75 - 375 + 1875 - 9375 , when r = - 5 .
則 S6 = 3 × ( 5 6 - 1 ) / (5-1) = 3/4 ×(5 6 - 1) or S6 = 3 × ( (-5) 6 - 1 ) / (-5-1) = -3/6 ×(5 6 - 1) = (1- 5 6)/2
0,0,0,0,....,0,0,0,0,... 為等差數列, 但非等比級數(因為公比為0/0 , 同時各項為0 )
1,1,1,1, ... , 1,1,... 為等差亦為等比數列
1,-1,1,-1,1,-1,..., (-1)n ,... 為正負交錯的交錯級數,也為等比級數
基礎集合理論
What is set ?
In mathematics, a set is a collection of distinct objects, considered as an object in its own right. 就是說為不同物件的收集. For example , 所有小於10 且大於0 的整數所成的集合
即 S = { 1,2,3,4,5,6,7,8,9 } . What is empty set (空集合) ? It means a set without any objects. It has a specified symbol ∅ .
What is subset ? (子集合)
一個集合中所有的元素皆為另一個集合的元素, 其間的關係以符號 ⊆ 表示
定義: ∅ 是任意集合的子集合.
Operations:
1. intersection (交集) : A ∩ B = { x | x ∈ A and x ∈ B }
2. Union (聯集) : A ∪ B = { x | x ∈ A or x ∈ B }
3. Difference (差集) : A - B = { x | x ∈ A , and x ∉ B }
Demorgan's Law - 迪摩根定律
1. ~ (A1∩A2∩A3∩A4∩ ... ∩An) = ~ A1∪ ~ A2 ∪ ~ A3 ∪ ... ∪ ~ An
2. ~ (A1∪A2∪A3∪A4∪ ...∪An) = ~ A1∩ ~ A2 ∩ ~ A3 ∩ ... ∩ ~ An
集合個數
假設有3個集合 A, B 及 C , 則n(A) 代表其集合的個數.
故 n(A ∪ B) = n(A) + n(B) - n(A ∩ B) , 再延伸3個集合的公式,
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) + n (A ∩ B ∩ C)
 |
| the union of three sets |
n(A1∪A2∪A3∪A4∪ ...∪An) = n(A1) + n(A2) + ... + n(An) - n(A1∩A2)- n(A1∩A2)- n(A1∩A2)
... - n(An-1∩An) + n(A1∩A2∩A3) + .... +(-1) n n(A1∩A2∩A3∩...∩An)
= ∑ n(Ai) - ∑ n(Ai ∩ Aj) + ∑ n(Ai∩Aj∩Ak) + ... + (-1) n n(A1∩A2∩A3∩...∩An) ,
討論其個數, 分析如下:
假設有n個集合, A1 , A2 , A3 , A4 , ... , An , 即取法如下:
n = 1 , nC1 = n , 有n個集合
n = 2 , nC1 = n(n-1) / 2 , i.e. 1+2+3+...+ n
n = 3 , nC3 = n(n-1)(n-2) / 6
...
n = n , nCn = 1
故其總和為 nC1 + (-1) nC2 + (-1) 2nC3 + (-1) 3 nC4 + ... + (-1) n-1 nCn , 考慮如下binomial expression
X1nC1 + X2nC2+ X3nC3 + ... + Xn nCn = (1+x)n
將令 x = - 1 , 如上的 expression 能取值如下 :
(-1)1nC1 + (-1)2nC2+ (-1)3nC3 + ... + (-1)n nCn 即所求的結果 .
∀n = 1,2,3,4, ...
Using mathematics induction to prove the above :
Induction Basis :
n = 1 , it is trivial .
Induction Hypotheses :
Assume n = m and m < n , p(n) is true , n(A1∪A2∪A3∪A4∪ ...∪Am) = ∑ n(Ai) - ∑ n(Ai ∩ Aj) + ∑ n(Ai∩Aj∩Ak) + ... + (-1) n n(A1∩A2∩A3∩...∩Am) is true .
Induction Steps :
Consider n = m +1 , n(A1∪A2∪A3∪A4∪ ...∪Am∪Am+1) = n( (A1∪A2∪A3∪A4∪ ...∪Am)∪Am+1)
= n(A1∪A2∪A3∪A4∪ ...∪Am) + n (Am+1) = ∑ n(Ai) - ∑ n(Ai ∩ Aj) + ∑ n(Ai∩Aj∩Ak) + ... + (-1) n n(A1∩A2∩A3∩...∩Am) + n (Am+1) -
n (Am+1 ∩ ( ∑ n(Ai) - ∑ n(Ai ∩ Aj) + ∑ n(Ai∩Aj∩Ak) + ... + (-1) n n(A1∩A2∩A3∩...∩Am) ) ) = ∑ n(Ai) - ∑ n(Ai ∩ Aj) +
∑ n(Ai∩Aj∩Ak) + ... + (-1) n n(A1∩A2∩A3∩...∩Am) + (-1) n+1 n(A1∩A2∩A3∩...∩Am+1) Q.E.D.
即如上的公式為排容原理 (Principle of Inclusion and Exclusion)
More information about this principle , please visit 國科會高瞻自然科學教學資源平台
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