AP Calculus BC Practice Exam 2 427
- The correct answer is (A).
∑∞
n= 1
n!+ 2 n
2 n·n!
=
∑∞
n= 1
n!
2 n·n!
+
∑∞
n= 1
2 n
2 n·n!
=
∑∞
n= 1
1
2 n
+
∑∞
n= 1
1
n!
The series
∑∞
n= 1
1
2 n
is a geometric series, so
∑∞
n= 1
1
2 n
=
1 / 2
1 − 1 / 2
=1.
Compare the series
∑∞
n= 1
1
n!
to the known
MacLaurin series
∑∞
n= 0
xn
n!
= 1 +x+
x^2
2!
+
x^3
3!
···=ex, and
atx=1,
∑∞
n= 0
xn
n!
=
∑∞
n= 0
1
n!
= 1 + 1 +
1
2!
+
1
3!
···=e^1. Thus,
∑∞
n= 1
1
n!
=e^1 −1.
(Note that the summation index changes from
n=0ton=1.) Therefore,
∑∞
n= 1
1
2 n
+
∑∞
n= 1
1
n!
= 1 +e− 1 =e. (Also, you
could have evaluated
∑∞
n= 1
1
n!
using your T1-89
calculator.)
- The correct answer is (B).
y=
1
x^2
−
1
x^3
=x−^2 −x−^3
⇒y′=− 2 x−^3 + 3 x−^4 ⇒y′′= 6 x−^4 − 12 x−^5. Set
the second derivative equal to zero and solve.
6
x^4
−
12
x^5
=
6 x− 12
x^5
= 0
⇒ 6 x− 12 = 0 ⇒x=2. Alsoy′′< 0
forx<2 andy′′>0 forx>2.
- The correct answer is (D).
The logistic growth model to the logistic
differential equation
dp
dt
=kp
(
1 −
p
M
)
with
Mequal to the maximum value is
p=
M
1 +Ae−kt
. In this case,p=
200
1 +Ae−.^25 t
and
t=0, you havep=20. Thus,
20 =
200
1 +Ae(−.25)(0)
or 20=
200
1 +A
orA=9,
Thus, atp=100, 100=
200
1 + 9 e−.^25 t
and
entering the equation into your calculator, you
havet≈ 8. 789 ≈9 years.