1(a) Consider the following sequence:
a1 = 1
a2= 1+2(1) = 3
a3= 1 + 2(1) + 2(2) =7
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a(n) =1+ 2(1) + 2(2) + ‧‧‧+ L
(i) Write down the values of a4 and a5. 我知道這條點做。
a4 = 1 + 2(1) + 2(2) + 2(3) = 13
a5 = 1 + 2(1) + 2(2) + 2(3) + 2(4) = 21
可否列式
(ii) Express the last term L of a(n) in terms of n, and hence simplify the
expression of a(n). Ans: L = 2 (n-1), a(n) = 1+ n(n-1)
(b) Consider the following sequence:
b1﹕1
b2﹕3, 5
b3﹕7, 9, 11
b4﹕ 13, 15,17,19
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b(n) ﹕a(n), a(n)+2, a(n)+4, ‧‧‧, L
(i) Express the last term L of b(n) in terns of n. Ans:L=n(n+1) -1=1
(ii) Using the result obtained in(a), or otherwise, find the sum of the terms in b(n).
Ans:n^3
2. An auditorium has 50 rows of seats. All seats are numbered in numerical
order from the first row to the last row, and from left to right, as shown in the
figure. The first row has 20 seats. The second row has 22 seat. Each
succeeding row has 2 more seats than the previous one.
The figure of the seats:
：
：
3rd row [ 43 | 44 | ‧‧‧‧‧‧65 | 66 ]
2nd row [ 21 | 22 |‧‧‧‧‧‧41 | 42 ]
1st row [ 1 | 2 |‧‧‧‧‧‧‧19 | 20 ]
(a) How many seats are there in the last row? 我知道這條點做。
a=20, d=2
T(50) = 20+ (50 -1)(2) = 118 seats
(b) Find the total number of seats inthe first n rows. 我知道 this ans is 19n+n^2
但是, 我不懂這條：
Hence determine in which row the seat numbered 2000 is located.
回答其中一條就可以。謝謝!
S.6 's math on AS and GS
2013-10-05 5:47 am
回答 (2)
2013-10-05 9:56 am
✔ 最佳答案
a1 = 1a2= 1+2(1) = 3
a3= 1 + 2(1) + 2(2) =7
a(1) = 1
a(2) = 1 + 2(1)
a(3) = 1 + 2(1) + 2(2)
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.
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a(n) =1+ 2(1) + 2(2) + ‧‧‧+ L
所以 L 是 2(n-1)
a(n) = 1 + 2[ 1 + 2 + 3 + ...... + (n-1) ]
注意 : 括號中共有 (n-1) 項，不是 n 項，代入 (首項 + 末項)X項數/2時要小心
= 1 + 2{[1+(n-1)](n-1)/2}
= 1 + (n-1) + (n-1)^2
= 1 + n-1 + n^2 - 2n + 1
= 1 + n^2 - n = 1 + n(n-1)
2013-10-05 02:01:53 補充：
b1 ﹕ [1] ................. 1個數
b2 ﹕ [3, 5] ..............2個數
b3 ﹕ [7, 9, 11]..........3個數
b4 ﹕ [13, 15,17,19]...4個數
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b(n) ﹕ [a(n), a(n)+2, a(n)+4, ‧‧‧, L] ...n個數
2013-10-05 02:08:32 補充：
b(n) ﹕ [a(n), a(n)+2, a(n)+4, a(n)+6, a(n)+8, a(n)+10, a(n)+12, ‧‧‧, L] ...n個數
...........1........2.........3.........4..........5..........6............7...........n
2013-10-05 02:11:46 補充：
注意 第 2 項 是 a(n)+2
第 3 項 是 a(n)+4......[ 4/2 + 1 = 3 ]
第 4 項 是 a(n)+6......[ 6/2 + 1 = 4 ]
第 5 項 是 a(n)+8......[ 8/2 + 1 = 5 ]
第 6 項 是 a(n)+10....[ 10/2 + 1 = 6 ]
第 n 項 是 a(n) + (n-1)X2
2013-10-05 02:17:06 補充：
由上面 1 a) ii) a(n) = 1+ n(n-1)
L = a(n) + (n-1)X2= 1+ n(n-1) + 2(n-1) = 1+n^2-n+2n-2 = n^2 + n - 1 = n(n+1) - 1
2013-10-05 02:22:40 補充：
S = a(n) + [a(n)+2] + [a(n)+4] + ‧‧‧+ [a(n) + 2(n-1)]
= [a(n)]n + 2 + 4 + 6 + 8 + ........ + 2(n-1)
計到這裏應該自己做了吧！
2013-10-06 4:55 am
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佮剸刲哷
佮剸刲哷
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