1. The positive integer 1, 2, 3, ..... are divided into groups G1, G2, G3, ..... , so
that the kth group G(k) consists of k consective integers as follows:
G1: 1
G2: 2, 3
G3: 4, 5, 6
.......
G(k - 1): u1, u2, ..... , u(k-1)
G(k): v1, v2, .... , v(k - 1), v(k)
.........
(a) (i) Write down all the integers in the 6th group G6.
( The ans is it 16, 17, 18, 19, 20, 21 ??)
我不懂得做以下問題, 可否教我
(ii) what is the total number of integers in the first 6 groups G1, G2, ... , G6?
(b) Find, in terms of k,
(i) the last integer u(k - 1) in G(k - 1) and the first integer v1 in G(k).
(ii) the sum of all the integers in G(k).
2. Suppose the number of babies born in HK in 1994 is 70000 and subsequent
years, the number of babies born each year increases by 2% of that of the
previous year.
(d) It is known that from 1901 to 2099, a year is a leap year閏年 if its
number is divisible by 4.
(i) Find the number of leap year between 1997 and 2046
(ii) Find the total number of babies born in HK in the leap years between 1997
and 2046
Math Q on AS &GS
2013-11-01 5:06 am
回答 (3)
2013-11-01 11:22 pm
✔ 最佳答案
第一題 Masterijk 博士已經解說得很清楚了，我就做第二題吧。The first leap year after year 1997 is 2000,
and the last leap year before year 2046 is 2044.
2000 + (n - 1)*4 = 2044
==> n = 12
(i) So, the number of leap year between 1997 and 2046 is 12.
(ii) Number of babies born in 2000 = 70000*1.02^6 ;
number of babies born in 2004 = 70000*1.02^10 ;
...
number of babies born in 2044 = 70000*1.02^50.
The above sequence is obviously a G.S. with 1st term = 70000*1.02^6,
common ratio = 1.02^4, number of terms = 12.
So, total number of babies born in these years is :
70000*1.02^6 * [(1.02^4)^12 - 1] / (1.02^4 - 1)
= 1518000 babies (corr to nearest thousands)
2013-11-01 8:44 am
wooooo.......i forgot them already:(:(:(:(:(
2013-11-01 6:40 am
加油呀! ☆ヾ(◕‿◕)ノ
1. (a) (ii)
Total number of integers in the first 6 groups G1, G2, ... , G6
= 1 + 2 + 3 + 4 + 5 + 6
= (1+6)*6/2
= 21
1. (b) (i)
Consider
u(k-1)
= 1 + 2 + 3 + ... + (k-1)
= [1+(k-1)]*(k-1)/2
= k*(k-1)/2
Then, v1 = u(k-1) + 1 = k*(k-1)/2 + 1
2013-10-31 22:40:57 補充：
1. (b) (ii)
The sum of all the integers in G(k) is
v1 + v2 + ... + v(k)
= [k*(k-1)/2 + 1] + [k*(k-1)/2 + 2] + ... + [k*(k-1)/2 + k]
= { [k*(k-1)/2 + 1] + [k*(k-1)/2 + k] } *k/2
= { k*(k-1) + k+1 } *k/2
= { k² - k + k + 1 } *k/2
= (k² + 1)k/2
1. (a) (ii)
Total number of integers in the first 6 groups G1, G2, ... , G6
= 1 + 2 + 3 + 4 + 5 + 6
= (1+6)*6/2
= 21
1. (b) (i)
Consider
u(k-1)
= 1 + 2 + 3 + ... + (k-1)
= [1+(k-1)]*(k-1)/2
= k*(k-1)/2
Then, v1 = u(k-1) + 1 = k*(k-1)/2 + 1
2013-10-31 22:40:57 補充：
1. (b) (ii)
The sum of all the integers in G(k) is
v1 + v2 + ... + v(k)
= [k*(k-1)/2 + 1] + [k*(k-1)/2 + 2] + ... + [k*(k-1)/2 + k]
= { [k*(k-1)/2 + 1] + [k*(k-1)/2 + k] } *k/2
= { k*(k-1) + k+1 } *k/2
= { k² - k + k + 1 } *k/2
= (k² + 1)k/2
收錄日期: 2021-04-13 19:46:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131031000051KK00204