mathematical induction

Let P be the proposition that "(3n+1)(7^n)-1 is divisible by 9" for any positive integers n.
when n=1, [3(1)+1](7)-1= 27, which is divisible by 9.
therefore, P is true when n=1.

Assume P is truw when n=k.
i.e. (3k+1)(7^k)-1 =9C, where C is a constant.
therefore, (3k+1)(7^k)-9C=1

when n=k+1,
[3(k+1)+1][7^(k+1)]-1
=[3k+4][7^(k+1)]-1
=[3k+4][7^(k+1)]-[(3k+1)(7^k)-9C]
=7^k(21k+28)-7^k(3k+1)+9C
=7^k(21k+28-3k-1)+9C
=7^k(18k+27)+9C
=7^k(9)(2k+3)+9C
=9[7^k(2k+3)+C], which is divisible by 9.

By MI, P is true for any positive integers n.

(3*1+1)(7^1) -1 = 27 is divisible by 9
assume (3k+1)(7^k) -1 is divisible by 9
(3k+1)(7^k) -1 = 9m where m is integer
consider
[3(k+1)+1][7^(k+1)] -1
= (3k+4)(7^k)(7) - 1
= (3k+1)(7^k)(7) + (3)(7^k)(7) - 1
= [(3k+1)(7^k) -1](7) + (21)(7^k) + 6
= 9m + (21)(7^k) + 6
如果証到 (21)(7^k) + 6 is divisible by 9,
就等於証到 (3n+1)(7^n) -1 is divisible by 9,
宜家變左要証另外一條 mathematical induction

又可以咁講 , 簡單 d,
如果証到 (7)(7^k) + 2 is divisible by 3
等於証到 (21)(7^k) + 6 is divisible by 9,

(7)(7^1) + 2 = 51 is divisible by 3
assume (7)(7^p) + 2 is divisible by 3
(7)(7^p) + 2 = 3q, q is integer
consider
(7)[7^(p+1)] + 2
= (7)(7^p)(7) + 2
= (7)[(7)(7^p) + 2] - 12
= (7)(3q) - 12
= 3(7q - 4)
7q - 4 is integer, so (7)[7^(p+1)] + 2 is divisible by 3
by M.I. , (7)(7^k) + 2 is divisible by 3
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2007-08-15 20:53:32 補充：
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