數學題目有些不會

Question

1.試求4的100次方除以9之餘數?
2.設k為整數若ｘ平方-kx+k+2=０之兩跟均為整數則ｋ＝？
3.設若複數平面上Ａ(2-i)B(4+3i)又三角形ABC之重心G(1+2i)則C點之對應複數為?
4.有一項等比數列前20項合為100,前40項合為2600,則前30項之和為?



請附算式

Answer 1

1. 4100 = 43*33+1= 6433*4
6433 = (9*7+1)33 = 1 (mod 9)
6433*4 = 1*4 = 4 (mod 9)
2. x2 - kx + k+2 = 0之兩跟均為整數
=> (-k)2 - 4(k+2) = c2 (c 為整數, c > 0)
=> k2 - 4k - (8+c2) = 0
=> k = 2 +- √(4+8+c2)
=> 設 4+8+c2 = p2 (p 為整數, p > 0)
=> p2 - c2 = 12
=> (p+c)(p-c) = 12
12 = 1*12 = 2*6 = 3*4
因為 p,c 為整數, p, c > 0
所以 p+c > p-c, p+c > 0
=> p+c = 12, p-c = 1 or p+c = 6, p-c = 2 or p+c = 4, p-c = 3
=> p=13/2 (不合), or p = 4, c =2, or p = 7/2 (不合)
=> k = 2 +- 4
=> k = 6 or -2
3. 若看成實數平面 則Ａ(2, -1), B(4, 3) ABC之重心G(1, 2)
重心G座標為 ( (Xa+Xb+Xc)/3, (Ya+Yb+Yc)/3)
=> 1 = (2+4+Xc)/3, 2 = (-1+3+Yc)/3
=> Xc = -3, Yc = 4
C點之對應複數為 -3+4i
4. 設首項 a 公比 r
則 S20 = 100 = a(1-r20)/(1-r)
S40 = 2600 = a(1-r40)/(1-r)
S40/S20 = 26 = (1-r40)/(1-r20)
令 r10 = A
=> 26(1-A2) = 1-A4
=> A4 - 26A2 + 25 = 0
=> (A2 - 25)(A2 - 1) = 0
=> A2 = 25 or 1 (不合)
=> A = 5
S30 = a(1-r30)/(1-r)
S30/S20 = S30/100 = (1-r30)/(1-r20)
= (1- A3)/(1 - A2)
= 124/24
=> S30 = 100*31/6 = 1550/3
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