a Calculate the number of books in the second and third year, then find a pattern.
B
b Use a graphing calculator.
A
aRecursive Rule: a_1=54 000, a_n=0.98a_(n-1)+1150 for n>1
B
b The number of books in the library stabilizes around 57 500 over time.
Practice makes perfect
a Let a_n denote the number of books in the library at the beginning of the n^(th) year. The library initially has 54 000 books. This tells us that a_1=54 000. Next, we will find a recursive equation for the sequence a_n.
Recursive Equation
At the beginning of the (n-1)^(th) year the number of books is a_(n-1). During this year, 2 % of the books are lost or discarded. Therefore, 98 %=0.98 of the old books remain. Furthermore, 1150 new books are added to the library.
ccccc
c Books at the start of the $n^(th)$ year
&
=0.98 *
&
c Books from previous year
&
+
&
New books
⇓ & & ⇓ & & ⇓
a_n & =0.98 * & a_(n-1) & + & 1150
Therefore, we can write a recursive rule: a_1=54 000, a_n=0.98a_(n-1)+1150 for n>1.
b We will graph the sequence from Part A using a calculator. First, we push MODE and change Func to Seq.
Now we are dealing with sequences in our calculator. Push Y=, and we get the following.
If we want to enter a recursive sequence we enter u(n-1) and u(n-2) for the previous terms. Enter 0.98*u(n-1) into the row with ⋱ u(n). The initial value or values we can enter in the row with u(nMin). Now, we will input the given recursive rule of the sequence from Part A into the calculator.
Now, to draw the sequence we push GRAPH.
By pressing the right arrow we can find that the values of the sequence stabilize around 57 500. This tells us that the number of books in the library stabilizes around 57 500 over time.
