Download quiz 5

1. Math 142 - Quiz 5 Solutions
(1) The Lucas numbers are a sequence of integers similar to the Fibonacci sequence. The Lucas
numbers Lk are defined as follows:
L1 = 2,
L2 = 1,
Lk = Lk−1 + Lk−2 .
Write the first six terms of the sequence Lk of Lucas numbers.
The recursive formula for Lk just says that each Lucas number is the sum of the previous
two Lucas numbers. So we have
L1 = 2
L3 = L2 + L1 = 1 + 2 = 3
L5 = L4 + L3 = 4 + 3 = 7
(2) Find the value of the sum
n
X
L2 = 1
L4 = L3 + L2 = 3 + 1 = 4
L6 = L5 + L4 = 7 + 4 = 11.
−5k + 2 as a function of n.
k=1
We have
n
X
k=1
−5k + 2 =
n
X
−5k +
k=1
= −5
n
X
k=1
k+
n
X
k=1
n
X
2
2
k=1
n(n + 1)
+ 2n
= −5
2
−5n2 − 5n
=
+ 2n
2
−5n2 − n
=
.
2
(MORE ON BACK)
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(3) A population of bateria grows at a constant (instantateous) rate of 10 individuals per hour
per 100 individuals. (I.e. the growth rate constant is 1/10.) I want to start with P cells,
then remove 300 cells after 1 hour (for an experiment) and still have P cells left in the
population. How large must my starting population of cells be?
Since the growth constant is 1/10, the population must follow the growth law
1
A(t) = P e 10 t
where P is the initial population. We know that the population after 1 hour must be 300
more than we started with (so that we can take out 300 and have P left over). So we have
1
P + 300 = A(1) = P e 10 ·1 .
Now we have an equation with only P in it, so we can solve for P :
1
P + 300 = P e 10
1
300 = P e 10 − P
1
300 = P (e 10 − 1)
300
P = 1
e 10 − 1
≈ 2852.499.
So we need to start with at least 2852 cells.