數學疑問 f(x) (20)

Question

題目:f(1)=2．f(2)=4． f(3)=6． f(x+3)=f(x)+5

．為分隔記號 非乘號
試問f(0)=?
f(976)
f(50)
f(51)
f(81)
如能有詳解更佳

f(2)=f(0+2)可以寫成f(0)+?嗎???
可以的話其值為?若不可以原因是?

Answer 1

 變數每增加三，函數值就增加5。所以變數每減少3，函數值就減少5。f(3)=f(0+3)=f(0)+5=6f(0)=1f(976)=f(973+3)=f(973)+5=f(976-3*1)+5*1=f(970+3*1)+5*1=f(970)+5*2=.....=f(976-3n)+5n976/3=325....1f(976)=f(976-3*325)+5*325=2+1625=1627f(50)=f(50-3*16)+5*16=f(2)+80=84f(51)=f(51-3*16)+5*16=f(3)+80=86f(81)=f(81-3*26)+5*26=6+130=136  

2006-06-07 21:13:45 補充：
f(976)=f(973+3)=f(973)+5=f(976-3*1)+5*1=f(970+3*2)+5*2=f(970)+5*2=.....=f(976-3n)+5n上面這個部份有打錯，更正為2。

2006-06-07 21:16:10 補充：
f(x)中的x如果為3n+1的型態，則可以化成f(1)+n*5，例如f(4)=f(3*1+1)=f(1)+5*1=7f(976)=f(3*325+1)=f(1)+5*325=1627

Answer 2

首先
f( 3 ) = f(0+3)
f(0+3) = f(0)+5
6 = f(0)+5
f(0)= 1

接著 f(976) f(50) f(51) f(81)
觀察以下數列規律 f(1)= 2
f(2)= 4
f(3)= 6
f(4)= f(1)+5 =7
f(5)= f(2)+5 =9
f(6)= f(3)+5 =11
f(7)= f(4)+5 =12
f(8)= f(5)+5 =14
f(9)= f(6)+5 =16


可得知
f(1+3n)=2+5n
f(2+3n)=4+5n
f(3+3n)=6+5n
〔n=大於0的整數〕


所以 f(976)= f(1+ 3 × 325 ) = 2 + 5 × 325 = 1627

f(50) = f(2+ 3 × 16 ) = 4 + 5 × 16 = 84

f(51) = f(3+ 3 ×16 ) = 6 + 5 × 16 = 86

f(81) = f(3+ 3 × 26 ) = 6 + 5 × 26 = 136

祝 學業進步

Answer 3

A:
f(1) = f(-2+3) = f(-2) + 5
f(2) = f(-1+3) = f(-1) + 5
f(3) = f( 0+3) = f( 0) + 5
由以上三式可推出
f(4) = f(1+3) = f(1) + 5 =2+5=7
f(5) = f(2+3) = f(2) + 5 = 4+5=9
f(6) = f(3+3) = f(3) +5 = 6+5=11

f(7) =f(4+3)=f(4)+5=f(1+3)+5=[f(1)+5]+5
.
.
.
以此可推
f(0+3)=f(0)+5
=>f(0)=f(0+3)-5=f(3)-5=1

f(976)=f(325*3+1)=f(1) + 325*5 = 2+ 325*5=1627
f(50)=f(16*3 + 2)=f(2) + 16*5 = 4+ 16*5 = 84
f(51)=f(16*3 + 3)=f(3) + 16*5 = 6+ 16*5 = 86
f(81)=f(26*3 + 3)=f(3) + 26*5 = 6+ 26*5 = 136

Answer 4

．為分隔記號 非乘號。呃！你可以用逗號，吧！類推，推導通式f(x+3)=f(x)+5f(x+6)=f(x+3)+5f(x+9)=f(x+6)+5.......f(x+3n)=f(x+3n-3)+5全部加起來，左右對消，得到f(x+3n)=f(x)+5nf(0)=f(0+3*1)-5*1=f(3)-5=6-5=1f(976)=f(1+3*325)=f(1)+5*325=2+1625=1627f(50)=f(2+3*16)=f(2)+5*16=4+80=84f(51)=f(0+3*17)=f(0)+5*17=1+85=86f(81)=f(0+3*27)=f(0)+5*27=1+135=136

2006-06-08 02:21:25 補充：
更一般的通式f(3n)=5n+1f(3n+1)=5n+2f(3n+2)=5n+4f(2)=f(0+2)可以寫成f(0)+?嗎???可以，由通式可知，f(3n+2)=f(3n)+3故f(2)=f(3*0+2)=f(3*0)+3=f(0)+3

Answer 5

f(0)=f(3)-5=1
f(976)=f(3*325+1)=2+325*5=1627
f(50)=f(3*16+2)=4+16*5=84
f(51)=f(3*16+3)=6+16*5=86
f(81)=f(3*26+3)=6+26*5=136