Answer:
![a) \ \ P_n=(1+r)P_{n-1}\\\\b)\ \ \ P_n=1.011^n(568000)\\\\c)\ \ 626,771\\\\d)\ \ 19.10yrs\ later \ (2027 \ February)](https://tex.z-dn.net/?f=a%29%20%5C%20%5C%20P_n%3D%281%2Br%29P_%7Bn-1%7D%5C%5C%5C%5Cb%29%5C%20%5C%20%5C%20P_n%3D1.011%5En%28568000%29%5C%5C%5C%5Cc%29%5C%20%5C%20626%2C771%5C%5C%5C%5Cd%29%5C%20%5C%2019.10yrs%5C%20later%20%5C%20%282027%20%5C%20February%29)
Step-by-step explanation:
a. Given that the population starts at 568000 and grows at 1.1% each years.
-The recursive formula for the population takes the form:
![P_n=(1+r)P_{n-1}](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29P_%7Bn-1%7D)
Where:
is the population at the nth year.
is the rate of growth
is the population a year before the nth year.
Hence, the recursive formula is given by ![P_n=(1+r)P_{n-1}](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29P_%7Bn-1%7D)
b. The explicit formula of a population growth takes the form:
![P_n=(1+r)^nP_o](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29%5EnP_o)
-Given that r=1.1% and the initial population is 568000
![P_n=(1+r)^nP_o\\\\r=1.1\%=0.011\\\\P_n=(1+0.011)^nP_o, P_o=568000\\\\P_n=1.011^n(568000)](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29%5EnP_o%5C%5C%5C%5Cr%3D1.1%5C%25%3D0.011%5C%5C%5C%5CP_n%3D%281%2B0.011%29%5EnP_o%2C%20P_o%3D568000%5C%5C%5C%5CP_n%3D1.011%5En%28568000%29)
Hence, the explicit formula is ![P_n=1.011^n(568000)](https://tex.z-dn.net/?f=P_n%3D1.011%5En%28568000%29)
c. The population in 2016 can be determined using the explicit formula.
-We substitute the growth rate and initial population as follows:
![P_n=(1+r)^nP_o\\\\=1.011^n(568000)\\\\n=2016-2007=9\\\\\therefore P_{2016}=(1+0.011)^9\times 568000\\\\=626,770.7721\approx626,771](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29%5EnP_o%5C%5C%5C%5C%3D1.011%5En%28568000%29%5C%5C%5C%5Cn%3D2016-2007%3D9%5C%5C%5C%5C%5Ctherefore%20P_%7B2016%7D%3D%281%2B0.011%29%5E9%5Ctimes%20568000%5C%5C%5C%5C%3D626%2C770.7721%5Capprox626%2C771)
Hence, the population in 2016 will be approximately 626,771
d. Given that the population after n years will be 700000.
#We substitute this value in the explicit formula to solve for n then add it to the initial year, 2007;
![P_n=(1+r)^nP_o\\\\700000=1.011^n(568000)\\\\1.011^n=\frac{700000}{568000}=\frac{700}{568}\\\\n=\frac{log \ (700/568)}{log\ 1.011}\\\\=19.10\ years\\\\\#Add \ 19.10yrs \ to \ 2007\\\\=2007+19.10yrs\\\\=2026.10\approx2027](https://tex.z-dn.net/?f=P_n%3D%281%2Br%29%5EnP_o%5C%5C%5C%5C700000%3D1.011%5En%28568000%29%5C%5C%5C%5C1.011%5En%3D%5Cfrac%7B700000%7D%7B568000%7D%3D%5Cfrac%7B700%7D%7B568%7D%5C%5C%5C%5Cn%3D%5Cfrac%7Blog%20%5C%20%28700%2F568%29%7D%7Blog%5C%201.011%7D%5C%5C%5C%5C%3D19.10%5C%20years%5C%5C%5C%5C%5C%23Add%20%5C%2019.10yrs%20%5C%20to%20%5C%202007%5C%5C%5C%5C%3D2007%2B19.10yrs%5C%5C%5C%5C%3D2026.10%5Capprox2027)
Hence, the population will get to 700000 after 19.10 years or in February the year 2027