123A-MATHS

Question

prove,by MI 3^2n-2^2n is divisble by 5

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Answer 1

最愛解答

3^(2(1)) - 2^(2(1)) = 5 is divisble by 5

assume 3^(2k) - 2^(2k) is divisble by 5

3^(2k) - 2^(2k) = 5m, where m is integer

consider

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

= (9)3^(2k) - (4)2^(2k)

= (9)[ 5m + 2^(2k) ] - (4)2^(2k)

= (5) 9m + (5)2^(2k)

= (5) (9m + 2^(2k) )

because 9m + 2^(2k) is integer, 3^(2(k+1)) - 2^(2(k+1)) is divisble by 5

so, when case n = k is true, case n = k+1 is also true

2007-11-03 19:41:32 補充：

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Answer 2

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Answer 3

以自我宣傳形式的最佳回答是不行的, 請接受公平競技

Answer 4

若能以公平的態度去對待兩位的回答則更好。

Answer 5

Let P ( n ) be the proposition “32n-22n is divisble by 5”.

When n = 1,

32-22=5 which is divisible by 5, so P ( 1 ) is true.

Assume P ( k ) is true for some positive integers k, i.e.

32k-22k=5M where M is an integer

When n = k + 1,

32k+2-22k+2

=9(32k)-4(22k)

=9(32k-22k)-4(22k)+9(22k)

=9(5M)+5(22k)

=5(9M+22k)

So P ( k + 1 ) is true.

By MI, P ( n ) is true for all positive integers n.

2007-11-05 18:44:00 補充：

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2008-06-26 17:58:48 補充：

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資料來源： My Maths Knowledge