maths. problem

1) 這是等差數列:

T ( n ) = a + ( n -1 ) d

T ( n ) = 3 + ( n - 1 )( 7 - 3 )

T ( n ) = 3 + 4n - 4

T ( n ) = 4n - 1

2) 這是等比數列:

T ( n ) = ar^ ( n -1 )

T ( n ) = ( 1 )( 4 / 1 )^( n - 1 )

T ( n ) = 4^ ( n - 1 )


2007-09-06 23:10:33 補充：
其實2^( 2n - 2 ) = 4^ ( n - 1 ), 因為4 ^ ( n - 1 )= ( 2^2 )^ ( n - 1 )= 2^[ 2 ( n - 1 )]= 2^( 2n - 2 ) 這只是表達方式的不同~~~

2007-09-06 23:14:35 補充：
再補充一點: 第一題的公差是4, 加上第一項是3這條件, 代入T ( n ) = a + ( n - 1 ) d , 便得T ( n ); 第二題的公比是4, 加上第一項是1這條件, 代入T ( n ) = ar^ ( n - 1 ), 便得T ( n )。

1. Tn = 4n -1
2. Tn = 2^(2n-1)
reasons:
1)
3, 7, 11, 15, 19, 23
= 4-1, 8-1, 12-1, 16-1, 20-1, 24-1
=4*1-1, 4*2-1, 4*3-1, 4*4-1, 4*5-1, 4*6-1
so you can see that it is 4*n -1 pattern

2)
1,4,16,64,256,1024
=2^0, 2^2, 2^4, 2^6, 2^8, 2^10
唔難見Tn係 2^一d野
咁睇個指數
0,2,4,6,8,10
第一項係0
通常係「n-一個數」
0,2,4,6,8,10
=2-2, 4-2, 6-2, 8-2, 10-2, 12-2
=2*1-2, 2*2-2, 2*3-2, 2*4-2, 2*5-2, 2*6-2
=2n-2 pattern
so Tn is 2^(2n-2)