# F1 mathematics-要快(要步驟)

1. The general term of a sequence is n+1 分之n-1.. Find a}the first five terms b}the 25th term of the sequence

2.In a hall,the number of2 seats n in a row is related to the row number L..It is known that n is afunction of L,and n =29+3L. Find the total number of seats in row 4 and row 5

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Chapter 4 P.182 Supplemtery EX 4

15a) The first 5 terms.

a(右下角n) = n-1/n+1

a(右下角1) = 1-1/1+1 = 0

a(右下角2) = 2-1/2+1 = 1/3

a(右下角3) = 3-1/3+1 = 2/4 = 1/2

a(右下角4) = 4-1/4+1 = 3/5

a(右下角5) = 5-1/5+1 = 4/6 = 1/3

15b) The 25th terms

a(右下角25) = 25-1/25+1 = 24/26 = 12/13

2007-11-26 22:30:54 補充：

17) The total number of seats in row 4.=29 3(4)=29 12=41∴It has 41 seats in row 4.The total number of seats in row 5.=29 3(5)=29 15=44∴It has 44 seats in row 5.The total number of seats in row 4 and 5.44 41=85∴It total has 85 seats.

2007-11-27 19:36:11 補充：

出吾到個加+號,,吾好意思系*****************************29 加 3(4)=29 加 12=41*********************************29 加 3(5) =29 加 15 =44*********************************44 加 41=85

Source(s): SySS人上
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• 1(a) n+1 分之n-1

1.(1)+1 分之(1)-1

=2分之0

=0

2.(2)+1分之(2)-1

=3分之1

=3

3.(3)+1分之(3)-1

=4分之2

=2

4.(4)+1分之(4)-1

=5分之3

1又2分之3

5.(5)+1分之(5)-1

=6分之4

=1又1分之2

1(b)(25)+1分之(25)-1

=26分之24

=1又1分之12

2.The total number of seats in row 4

=29+3(4)

=29+12

=41 seats

The total number of seats in row 5

=29+3(5)

=29+15

=44 seats

2007-11-26 20:18:55 補充：

1.(a)2.(2)+1分之(2)-1=3分之13.(3)+1分之(3)-1=4分之2=2分之14.(4)+1分之(4)-1=5分之35.(5)+1分之(5)-1=6分之4=3分之2

2007-11-26 20:19:31 補充：

1(b)(25) 1分之(25)-1=26分之24=13分之12

Source(s): me
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• 1a. The first five terms are order

T(1)=0, T(2)=1/3, T(3)=1/2, T(4)=3/5, T(5)=2/3

b. T(25)=24/26=12/13

2. Row 4=29+3(4)=41

Row 5=29+3(5)=44

Source(s): me
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