1.Mr. Cheung saves his money by despositing 1000 dollars in a bank at the beginning of each year.The interest rate is fixed at 4 percent p.a.,compounded yearly.What is the minimum number of years needed to obtain an amount greater than 15000 dollars？
2.How many terms of each of the following geometric series must be taken to give the sum stated？
-1/3+1-3+9-...=182/3
Maths Question 急
2007-11-16 5:40 am
回答 (2)
2007-11-16 7:21 am
✔ 最佳答案
(1)That is G.P.
a = 1000, r = 1.04
a + ar + ar^2 + ... + ar^(n-1) > 15000
1000 + 1000 x 1.04 + 1000 x 1.04^2 + ... +1000 x 1.04^(n-1) > 15000
1000 [1.04^n - 1] / (1.04 - 1) > 15000
1.04^n - 1 > 0.6
1.04^n > 1.6
n log 1.04 > log 1.6
n > 11.98
n = 12 years
(2)
a = -1/3, r = -3
a + ar + ar^2 + ... +ar^(n-1) = 182/3
a(r^n - 1) / (r-1) =182/3
(-1/3) [(-3)^n - 1] / (-3-1) = 182/3
(-3)^n - 1 = 728
(-3)^n = 729
n is an even number
(3)^n = 729
n log 3 = log 729
n = 6
參考: My calculation
2007-11-16 7:37 am
1. 1000(1.04)^n >15000
log1000 + n log(1.04) > log(15000)
n> (log15000 -log1000) / log(1.04)
n>69.046
n>70
2. Sn= a(1-r^n) / 1-r
a=-1/3 , r =-3
-1/3 [1-(-3)^n] /(1+3) =182/3
1-(-3)^n=182 (4)(-3)/3
1-(-3)^n=-728
729 = (-3)^n
n=6
log1000 + n log(1.04) > log(15000)
n> (log15000 -log1000) / log(1.04)
n>69.046
n>70
2. Sn= a(1-r^n) / 1-r
a=-1/3 , r =-3
-1/3 [1-(-3)^n] /(1+3) =182/3
1-(-3)^n=182 (4)(-3)/3
1-(-3)^n=-728
729 = (-3)^n
n=6
收錄日期: 2021-04-24 09:50:22
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https://hk.answers.yahoo.com/question/index?qid=20071115000051KK03864