2條A.MATHS(m.i)(快)

1)Let P ( n ) be the proposition
“( 3n + 1 )7n – 1 is divisible by 9”.
When n = 1,
( 3 + 1 )( 7 ) – 1
= 27 which is divisible by 9, so P ( 1 ) is true.
Assume P ( k ) is true for some positive integers k, i.e.
( 3k + 1 )( 7k ) – 1 = 9M where M is an integer
When n = k + 1,
( 3k + 3 + 1 )( 7k+1 ) – 1
= 7 ( 7k )( 3k + 1 ) + 7 ( 7k )( 3 ) – 1
= 7 [( 7k )( 3k + 1 ) – 1 ] + 7 ( 7k )( 3 ) + 6
= 7 ( 9M ) + 21 ( 7k )+ 6
= = = = = =
Then let P1 ( n ) be the proposition
“21 ( 7n )+ 6 is divisible by 9”.
When n = 1,
21 ( 7 ) + 6 = 153 which is divisible by 9
So P1 ( 1 ) is true.
Assume P1 ( k ) is true for some positive integers k, i.e. 21 ( 7k )+ 6= 9N where N is an integer
When n = k + 1,
21 ( 7k+1 )+ 6
= 21 ( 7 )( 7k ) + 6
= 7 [ 21 ( 7k ) + 6 ] – 36
= 7 ( 9N ) – 36
= 9 ( 7N – 4 )
So P1 ( k + 1 ) is true. By MI, P1 ( n ) is true for all positive integers n.
= = = = = =
Then 7 ( 9M ) + 21 ( 7k )+ 6
= 7 ( 9M ) + 9 ( 7N – 4 )
= 9 ( 7M + 7N – 4 )
So P ( k + 1 ) is true. By MI, P ( n ) is true for all positive integers n.
2)Let P ( n ) be the proposition “ xn + yn is divisible by x + y for all positive odd integers n”.
When n = 1,
x1 + y1 = x + y which is divisible by x + y, so P ( 1 ) is true.
Assume P ( k ) is true for some positive odd integers k, i.e.

xk + yk = M ( x + y ) where M is an integer
When n = k + 2,
xk+2 + yk+2
= x2 ( xk ) + y2 ( yk )
= x ( xk + yk ) – x2 ( yk ) + y2 ( yk )
= Mx ( x + y ) – ( yk )( x2 – y2 )
= Mx ( x + y ) – ( yk )( x + y )( x – y )
= ( x + y )[ Mx – ( yk )( x – y ) ]
So P ( k + 1 ) is true.
By MI, P ( n ) is true for all positive odd integers n.



2008-07-06 16:02:47 補充：
Sorry, 尾二嗰行打錯咗:
So P ( k + 2 ) is true.

Let P ( n ) be the proposition

“( 3n + 1 )7n – 1 is divisible by 9”.