Calculate the few first terms of the sequence that describe the number of fish in the lake.
Will the lake run out fish? No, the population of the lake converges to 12 500. See solution for the explanation.
Practice makes perfect
Let a_n be the number of fish in the lake after the n^(th) year. The initial population is a_0= 10 000. Let's find a formula for a_n for n≥ 1. At the beginning of this year we have a_(n-1) fish. Since 80 % of the fish die every year, only 20 %= 0.2 remain, 0.2 a_(n-1). Furthermore, the lake is replenished with 10 000 new fish.
a_n= 0.2 a_(n-1)+ 10 000forn≥ 1
We will also round the value of a_n to the nearest integer. Now, let's calculate the few first terms of the sequence a_n using the rule a_n= 0.2 a_(n-1)+ 10 000 for n≥ 1.
n
a_n=0.2a_(n-1)+10 000
0.2a_(n-1)+10 000
a_n
0
a_0=10 000
-
a_0= 10 000
1
0.2a_0+10 000
0.2( 10 000)+10 000= 12 000
a_1= 12 000
2
0.2a_1+10 000
0.2( 12 000)+10 000= 12 400
a_2= 12 400
3
0.2a_2+10 000
0.2( 12 400)+10 000=12 480
a_3=12 480
4
0.2a_3+10 000
0.2(12 480)+10 000=12 496
a_4=12 496
5
0.2a_4+10 000
0.2(12 496)+10 000= 12499.2
a_5= 12 499
6
0.2a_5+10 000
0.2( 12 499)+10 000= 12 499.96
a_6= 12 500
7
0.2a_6+10 000
0.2( 12 500)+10 000= 12 500
a_7= 12 500
The lake will not run out fish, because there will always be 20 % of population from the last year. Let's notice that a_6=a_7=12 500. This tells us that after the 6^(th) year the number at the end of each year will be constant and equal to 12 500. Therefore, the number of fish converges to 12 500.
Extra
Calculations Using a Spreadsheet
Let's enter a_1=10 000 in cell A1 and write =Round((0.2)*(A1)+10000,0) in cell A2. When we hit enter, we will get the following.
Next, copy cell A2, highlight cells A3 through A10, and paste.
From the spreadsheet we can see that the values of the sequence stabilize around 12 500.
