geometric series ...

The population of a city increases in such a way that if it is p at the beginning of a year , then at the end of that year it is p X 1.01+3000. Given that the population of the city is 500 000 at the beginning of the first year.
a)Find the population of the city at the end of
i )the first year .
500,000x1.01 + 3000
= 508,000

ii )the second year.
508,000x1.01 + 3000
= 516,080

b)Show that the population at the end of the nth year is 800 000(1.01)n - 300 000.
at the end of the n th year
500,000x1.01n + 3000 + 3000x1.01 + 3000x1.012 + ….. + 3000x1.01n-1
使用等比級數和的公式
S = (arn – a) / (r – 1)
a = 3000；r = 1.01
所以
3000 + 3000x1.01 + 3000x1.012 + ….. + 3000x1.01n-1
= (3,000x1.01n – 3000) / (1.01 – 1)
= 300,000x1.01n - 300000
500,000x1.01n + 3000 + 3000x1.01 + 3000x1.012 + ….. + 3000x1.01n-1
= 500,000x1.01n + 300,000x1.01n – 300,000
= 800,000x1.01n – 300,000

c)Find,correct to 3 significant figures, the population of this city at the end of the tenth
十年後的本利和
= 800,000x1.0110 – 300,000
= 583697.7
= 584000(三位有效數字)

answer:(a) (i)508 000 (ii)516 080 (b)584 000