In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.2: Exponential Functions
Problem 54E
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In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?

Example 8.2.3: Compound Interest
Suppose you are offered two retirement savings plans. In Plan A, you start with
$1,000, and each year (on the anniversary of the plan), you are paid 11% simple
interest, and you add $1,000. In Plan B, you start with $100, and each month, you
are paid one-twelfth of 10% simple (annual) interest, and you add $100. Which plan
will be larger after 40 years?
// Can we apply a recurrence equation?
Consider Plan A and let S, denote the number of dollars in the plan after
(exactly) n years of operation. Then So
$1,000 and
Sn+1 = Sn + interest on S,+$1000
= S, +11% of S,
= S,(1+0.11)
+ $1000
+ $1000.
1000
In this RE, a = 1.11, c=
1000, so
and
1 - a
-0.11
1000
S, = (1.11)" | 1000
1000
+
-0.11
-0.11
1110
1000
= (1.11)"
+0.11
+0.11
Transcribed Image Text:Example 8.2.3: Compound Interest Suppose you are offered two retirement savings plans. In Plan A, you start with $1,000, and each year (on the anniversary of the plan), you are paid 11% simple interest, and you add $1,000. In Plan B, you start with $100, and each month, you are paid one-twelfth of 10% simple (annual) interest, and you add $100. Which plan will be larger after 40 years? // Can we apply a recurrence equation? Consider Plan A and let S, denote the number of dollars in the plan after (exactly) n years of operation. Then So $1,000 and Sn+1 = Sn + interest on S,+$1000 = S, +11% of S, = S,(1+0.11) + $1000 + $1000. 1000 In this RE, a = 1.11, c= 1000, so and 1 - a -0.11 1000 S, = (1.11)" | 1000 1000 + -0.11 -0.11 1110 1000 = (1.11)" +0.11 +0.11
8.2 Solving First-Order Linear Recurrence Equations
345
Hence, S40 = (1.11)"(10 090.090 909...)
-(9 090.909 090. ..)
- (9 090.909 090...)
(65.000 867...)(10 090.090 909...)
655 917.842. ..
- (9 090.909 090. ..)
e $646 826.
// Can that be right? You put in $40,000 and take out > $600,000 in interest.
Now consider Plan B and let T, denote the number of dollars in the plan after
(exactly) n months of operation. Then To
$100 and
Tn+1 = Tn+interest on T,
+$100
= T,+(1/12) of 10% of T, + $100
= T„[1+0.1/12]
+$100.
100
In this RE, a = 12.1/12,
= 100, so
-12000 and
-0.1/12
T, = (12.1/12)"[100+ 12000] – 12000.
1 - a
Hence, after 40 × 12 months,
(12.1/12)*8° (12100)
: (1.008 333...)*8" (12100) - (12000)
(53.700 663...)(12100)
649 778.023 4...
480
T480
- (12000)
480
– (12000)
- (12000)
%3|
e $637 778.
Therefore, Plan A has a slightly larger value after 40 years.
Transcribed Image Text:8.2 Solving First-Order Linear Recurrence Equations 345 Hence, S40 = (1.11)"(10 090.090 909...) -(9 090.909 090. ..) - (9 090.909 090...) (65.000 867...)(10 090.090 909...) 655 917.842. .. - (9 090.909 090. ..) e $646 826. // Can that be right? You put in $40,000 and take out > $600,000 in interest. Now consider Plan B and let T, denote the number of dollars in the plan after (exactly) n months of operation. Then To $100 and Tn+1 = Tn+interest on T, +$100 = T,+(1/12) of 10% of T, + $100 = T„[1+0.1/12] +$100. 100 In this RE, a = 12.1/12, = 100, so -12000 and -0.1/12 T, = (12.1/12)"[100+ 12000] – 12000. 1 - a Hence, after 40 × 12 months, (12.1/12)*8° (12100) : (1.008 333...)*8" (12100) - (12000) (53.700 663...)(12100) 649 778.023 4... 480 T480 - (12000) 480 – (12000) - (12000) %3| e $637 778. Therefore, Plan A has a slightly larger value after 40 years.
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