教學指標 (11)

盤算數學 (11)
代數原本,應用延伸。
 ❶E  =  A  +1  ,  E後項 , A前項    
    0,1,2,3,4,5,6,7,8,9 ,,,
整數一個一數
連續整數的代數基本式
  ∵ E = A +1   ∴ E - 1 = A  倒數
一段算盤的代數式
   E  =  A  +1 , E進位數,A各檔珠數
二段算盤, E進位數,AB各檔珠數
   E  =  (A+1)×(B+1)
三段 , E進位值 , ABC各檔珠數
   E  =  (A+1)×(B+1)×(C+1)
四段 , E進位數,ABCD各檔珠數
   E  =  (A+1)×(B+1)×(C+1)×(D+1)
算盤  段數≧1, 檔數≧1, 珠數≧1
數線上點數與間隔數  E≧2
   E  =  A  +1 , E點數,A間隔數
  ❹E  =  A  幾何圖形的代數式
   E 頂點數 , A 邊數 ,   E≧3
連續整數自首A數至末E有N個
   A , A+1 , A+2 , A+3 , , , E
   E > A  ,    N    =  E  - A  +1
                 個數 = 大 - 小 +1
   2 個一數自 A 數至 E 有N個
   A , A+2 , A+4 , A+6 , , , E
          N = (E - A) ÷2 +1
   3 個一數自 A 數至 E 有N個
   A , A+3 , A+6 , A+9 , , , E
          N = (E - A) ÷3 +1
   d 個一數自 A 數至 E 有N個
   A , A+d , A+2d , A+3d , , , E
          N = (E - A) ÷d +1
等差數列的差數d,即 d 個一數
個數 = (末數 - 首數) ÷差數 + 1
末數 = (個數  - 1 ) ×差數 +首數
等差數列的差數d,首數A,個數N
求末數E?
   ∵  N = (E - A) ÷d +1
   ∴ (N - 1) ×d +A = E
☆移項原理如同 : 一步一腳印
    圖示 : 歸零  手撥珠,眼看,腦驗證

Computing Mathematics (11)
Algebraic original, application extension.
 ❶E = A +1, E after the item, A before the item
    0,1,2,3,4,5,6,7,8,9 ,,,
The integer is one by one
The Algebraic Basic of Continuous Integer
  ∵ E = A +1 ∴ E - 1 = A reciprocal
An almanac of an abacus
   E = A +1, E carry digits, A number of stalls
Two abacus, E carry the number of, AB the number of stalls
   E = (A + 1) × (B + 1)
Three paragraphs, E carry value, ABC file number of beads
   E = (A + 1) × (B + 1) × (C + 1)
Four, E carry the number of ABCD file number
   E = (A + 1) × (B + 1) × (C + 1) × (D + 1)
Abacus number ≧ 1, the number of stalls ≧ 1, beads ≧ 1
Number of points on the line and the number of intervals E ≧ 2
   E = A +1, E points, A number of intervals
  Algebraic representation of geometric figures
E vertex number, A side number, E ≧ 3
Continuous integer surrendered A number to the end of E have N
   A, A + 1, A + 2, A + 3,,, E
   E> A, N = E - A +1
Number = big - small +1
   2 number from A number to E have N number
   A, A + 2, A + 4, A + 6,,, E
          N = (E - A) ÷ 2 +1
   3 number from A number to E have N number
   A, A + 3, A + 6, A + 9,
          N = (E - A) ÷ 3 +1
   D number from A number to E have N number
   A, A + d, A + 2d, A + 3d,
          N = (E - A) ÷ d +1
The difference d of the difference series is the number of d
Number = (the last number - the first number) ÷ difference + 1
The last number = (number - 1) × difference number + first number
The difference number d, the first number A, the number N of the difference series
The final number of E
   ∵ N = (E - A) ÷ d +1
   ∴ (N - 1) × d + A = E
☆ shift as the principle:
step by step footprints
Illustration: Zero hand dial, seeing, brain verification