Induction

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2010-04-20 22:36:55 補充：
Because k and A are postive integers.

For n = 1 :

4 + 2*3 + 2 = 12 is true,

Let n = k

i.e 4^k + 2(3^k) + 2 is divisible by 12 be true , let it be 12S :

when n = k+1 :

4^(k+1) + 2(3^(k+1)) + 2

= 4(4^k) + 6(3^k) + 2

= 4(4^k) + 8(3^k) + 8 - (2(3^k) + 6)

= 4(12S) - (2(3^k) + 6)

2010-04-19 21:00:02 補充：
= 4(12S) - [2(3^k) + 3(12S - 4^k - 2(3^k))].....Since 2 = 12S - 4^k - 2(3^k)

= 4(12S) - [ - 4(3^k) + 3(12S - 4^k) ]

= 4(12S) - 4[ - (3^k) + 3(3S - 4^(k-1)) ]

= 4(12S) - 12[ - (3^(k-1) + (3S - 4^(k-1)) ]

= 12 { 4S - [ - (3^(k-1) + (3S - 4^(k-1)) ] }

which is divisible by 12.