Question

Assume that the population of the world in 2017 was 7.6 billion and is growing at the rate of 1.12% a year. a) Set up a recurrence relation for the population of the Links world n years after 2017. b) Find an explicit formula for the population of the world n years after 2017. c) What will the population of the world be in 2050?

Answer 1

Answer:

a)         $= 1.0112 a_{n-1}$

b)         $= 7.6 . 1.0112^n$

c) $a_{33} = 10.97 billion$

Step-by-step explanation:

Given data:

Population in 2017 was 7.6 billion

r = 1.12%

a) population after n year 2017

It is given each year population rise at a rate of 1.12% Thus we have

$a_n = a_{n-1} + 1.12%.a_{n-1}$

       $= a_{n-1} + 0.0112 a_{n-1}$

       $= 1.0112 a_{n-1}$

b) $a_{n} = 1.0112 a_{n-1}$

   $a_{n} = 1.0112 a_{n-1} = 1.011 ^1 a_{n-1}$

   $a_{n} = 1.0112 a_{n-2} = 1.011 ^2 a_{n-2}$

   $a_{n} = 1.0112 a_{n-3} = 1.011 ^3 a_{n-3}$

      ....

     $= 1.0112^n a_{n-n}$

      $= 1.0112^n a_{0}$

      $= 7.6 . 1.0112^n$

c)  for n = 33 year ( 2050- 2017 = 33 year)

        $a_{33} = 7.6 \times 1.0112^33$

$a_{33} = 10.97 billion$

Answer 2

Answer:

Answered

Step-by-step explanation:

  $a_0= 736 billion$

    r=    1.12%=   0.0112

a) let a_n represents population n years after 2017

each year population grows by 1.12 %. Thus the population is the population of the previous year multiplied by a factor of 1.12%.

that is

$a_n =a_{n-1} +1.0112a_{n-1}$

$a_n =1.0112a_{n-1}$

b) given $a_n =1.0112a_{n-1}$

$a_0= 736 billion$

we successively apply the recurrence relation:

a_n= 1.0112a_n-1 = 1.0112^1a_n-1

$1.0112(1.0112a^{n-2})= 1.0112^2 a_{n-2}$

$1.0112^2(1.0112a^{n-3})= 1.0112^3 a_{n-3}$

$1.0112^3(1.0112a^{n-4})= 1.0112^4 a_{n-4}$

.......................

=1.0112^na_n-n

=7.6×1.0112^n

c) the population of the world be in 2050

n=33 years

=7.6×1.0112^33

=10.975 billion