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||Définition d'une variable || Algèbre || Arithmetique || Logique ||La raison basée sur les règles et modelés||
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YOU are better than YOU think. Show yourself  how:  

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Read  logic chapters 1 to 5  in online volume Three Skills for Algebra  for greater skills & confidence in  work 
and study.

Learn to read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention.

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 Logic chapters 1 to 5  re- appear not in sequence, as is or longer,  in  Volume 1A,  Pattern Based Reason, Bon Appetite.

Logic Mastery
 Amazing, Amusing, Amorous,  Delicious, Delightful, Edifying, Strengthening Elixir. 
It eases work & learning difficulties Makes the hard easier. Opens eyes. Leads to greater precision.
in reading and
writing

Logic mastery makes the hard, easier. Logic mastery  leads to better, stronger and richer comprehension.  Logic mastery  improves reading and writing.  Logic mastery ease learning difficulties.  Logic mastery gives a headstart.  In sum, logic mastery  will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck.


After logic  (a) continue reading Three Skills for Algebra, chapters 8 to 14  and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes  & More Math, chapters 2 to 6;

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Caution: Site advice is approximately correct, for some circumstances, not all. That leaves room for thought

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What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts.


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For online automated help in senior high school maths & calculus, visit  quickmath.com  For Automatic Calculus and Algebra Help with derivatives, integrals, graphs, linear equations, matrix algebra, visit calc101.com  With  overlap, each site quickmath & calc101offers a different range of services, some free, some not, all based on webmathematica. Good luck.

Chapter 24
Personal Money, Investment
and Pension Computations

Previous Section: Periodic Deposits; Credit, Loans, Payments; Account Tracking

4  Pension Plan Example

Example:   John is seventy years old. The A1Z26 Pension Provider Company offers (sells) a 15 year annuity pension plan consisting of constant monthly payments. The size of the payment of the payment depends on the cost of the plan and present interest rate assumptions. John has up to $350000 dollars to buy a plan. The A1Z26 Pension Provider company announces an interest rate of 4 percent, compounded monthly. What is the maximum monthly payment2 John can obtain?

First Solution: First imagine that the $350000 dollars is placed in a credit account with the company, also at 4 percent, compounded monthly. Then this credit grows to the amount
Acredit = $350000·(1+ .04
12
)180
Therefore
Acredit = $350000·(1.82030163) = $637105.57
This amount could be paid to the company at the end of the fifteen years.

Second, imagine that John has asked the A1Z26 Pension Provider Company to let him debit $1.00 from his pension plan the end of each month for the next 15 years, at the interest rate of 4 percent compounded monthly. We will calculate the amount of debt that the payments represent at the end of fifteen years, and then see what amount of credit is then required to cancel this debt.

    1. At the end of the first month, the first payment of $1.00 to John represents a debt which grows at 4 percent, compounded monthly for 180-1 =179 months. This contributes a debt of $1.00(1+[.04/12])179 = (1+[.04/12])179 dollars.
    2. At the end of the second month, the second payment of $1.00 to John represents a debt which grows at 4 percent, compounded monthly for 180-2 =178 months. This contributes a debt of $1.00(1+[.04/12])178 = (1+[.04/12])178 dollars.
    3. At the end of the third month, the third payment of $1.00 to John represents a debt which grows at 4 percent, compounded monthly for 180-3 =177 months. This contributes a debt of $1.00(1+[.04/12])177 = (1+[.04/12])177 dollars.
    .
    .
    .

    178. At the end of the 178th month, the 178th payment of $1.00 to John represents a debt which grows at 4 percent, compounded monthly for 180-178 =2 months. This contributes a debt of $1.00(1+[.04/12])2 = (1+[.04/12])2 dollars.
    179. At the end of the 179th month, the 179 payment of $1.00 to John represents a debt which grows at 4 percent, compounded monthly for 180-179 =1 month. This contributes a debt of $1.00(1+[.04/12])1 = (1+[.04/12])1 dollars.
    180. At the end of the 180th month, the last payment of $1.00 to John or his estate represents a debt which grows at 4 percent, compounded monthly for 180-180 =0 months. This contributes a debt of $1.00(1+[.04/12])0 = 1 dollars.
The total amount of debt at the end of fifteen years is given by the geometric sum
Adebt = $ é
ê
ë
(1+ .04
12
)179+(1+ .04
12
)178+ (1+ .04
12
)177+
   +¼+ (1+ .04
12
)2+ (1+ .04
12
)1+ (1+ .04
12
)0 ù
ú
û
=
$ 179
å
j = 0 
(1+ .04
12
)j
The value of this geometric sum is
Adebt = $
(1+ .04
12
)180-1

1+ .04
12
-1
Therefore
Adebt = $

1.820301518-1
.04
12
= $246.09
Thus 180 end of month payments of $1.00 from the A1Z26 Pension Provider Company will result in a debt of 246.09 to the A1Z26 Pension Provider Company at the end of the fifteen years.

Now at the end of fifteen years, a deposit of 3,500,000 dollars today will leave John with a credit of 637105.57 dollars. Observe
637,105.57¸246.09 = 2588.91
Therefore if John receives 2588.91 dollars at the end of each month for the next 15 years = 180 months, he will need to give 637,105.57 = 2588.91×246.09 dollars to the A1Z26 Pension Provider Company at the end of the fifteen years. But that is precisely the accumulated amount in his credit account.

This suggests that the A1Z26 Pension Plan Company would be willing to sell John fifteen years of monthly payments of 2588.91 dollars (approximately) in exchange for the amount of 350000 dollars at their announced 4% cent interest rate.

Second Solution:   Suppose John want to receives a payment of 1.00 dollar at the end of each month for the next fifteen years = 180 months. Recall that a deposit of P = [(A)/((1+i)n)] will grow the amount A at the end of n periods in a compound interest account. Suppose John want to receives a payment of 1.00 dollar at the end of each month for the next fifteen years = 180 months from a compound interest account, paying 4 percent compounded monthly. What does he need today in the account today to be able to make all the withdrawals?


    1. To withdraw 1.00 dollar from the account one month after his initial deposit, he needs [$1.00/((1+[.04/12])1)] in the initial deposit.
    2. To withdraw 1.00 dollar from the account two months after his initial deposit, he needs [$1.00/((1+[.04/12])2)] in the initial deposit.
    3. To withdraw 1.00 dollar from the account three months after his initial deposit, he needs [$1.00/((1+[.04/12])3)] in the initial deposit.
    .
    .
    .

    177.
    To withdraw 1.00 dollar from the account 178 months after his initial deposit, he needs [$1.00/((1+[.04/12])178)] in the initial deposit.
    179. To withdraw 1.00 dollar from the account 179 months after his initial deposit, he needs [$1.00/((1+[.04/12])179)] in the initial deposit.
    180. To withdraw 1.00 dollar from the account 180 months after his initial deposit, he needs [$1.00/((1+[.04/12])180)] in the initial deposit.
To make all the withdrawals the total amount in the initial deposit has to be
Adeposit =
$1.00
(1+ .04
12
)1
+

 


$1.00
(1+ .04
12
)2
+

 


$1.00
(1+ .04
12
)3
+ ¼

   + 


$1.00
(1+ .04
1
)178
+

 


$1.00
(1+ .04
12
)179
+

 


$1.00
(1+ .04
12
)180

Therefore
Adeposit =

$1.00
(1+ .04
12
)180
é
ê
ë
(1+ .04
12
)179 + (1+ .04
12
)178
   +(1+
.04
12
)177+¼+(1+
.04
12
)2+(1+
.04
12
)1+1 ö
÷
ø

From the value of the geometric sum in the last expression, we conclude
Adeposit =



$1.00
(1+
.04
12
)180
·
(1+
.04
12
)180-1

1+
.04
12
-1
Therefore
Adeposit =



$1.00
1.820301518



·


1.820301518-1
.04
12
= $135.1921388

Therefore, for each dollar John receives at the end of each month for the next 15 years = 180 months, he will need to place an initial deposit of 135.1921388 dollars with the A1Z26 Pension Provider Company. Now 350000¸138.1921388 = 2588.91. Therefore for John to receive 2588.94 dollars at the end of each month for the next 15 years = 180 months, he will need to make an initial deposit of 350000 = 2588.91×135.19 dollars to the A1Z26 Pension Provider Company. This assumes the 4% yearly interest rate, compounded monthly.

Note in computing, rounding should be done last and avoided in intermediate calculations for the sake of preserving accuracy. (The above calculation would be slightly less accurate if the intermediate result 135.192188 was rounded to 135.19 dollars.)


More Chapter Sections: Up ] 24 Periodic Deposits ] 24 Account Tracking ] [ 24 Pension Plans ]

Next Chapter: 25 Inductive Proofs and Recursive Definitions

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2. Three Skills for Algebra 

Foreword, Chapters 
& Appendices 

Foreword
1. Introduction
2. Implication Rules
3. Chains of Reason
4. Romeo and Juliet
4. Induction Mathematical
5 Knowledge Islands
6  Old Language
7  Arith Skill Check
7. The Next Chapters
8 The Three Skills
8 VNR-Concise-Encyclopedia
PS. What is a Variable
9. Algebra Talk
10 Two More Skills
11 Why Shorthand
12 Shorthand Usage
13 What's Next
14 Compound Interest
15 Linear Equations
PS I.  Distributive Law
PS II. Polynomials
16 Painless Proofs
17 Pythagoras
18 Rules of Algebra
19  Functions & Sets
20 Degrees & Radians
21 What's Next
22. Arith & Geometric Sums
23 Summation Notation
24 Your Money
25 Induction & Recursion
26 What's Next
27 Pronouns in Logic
28 Occurrence Tables
29 Contrapositive
30 Truth Tables
31 Indirect Reason
A. Advice For Learning

Real Player Videos

Perfect arithmetic skills with whole numbers & fractions
after or besides chapters 1 to 14.

Arithmetic Videos Summary
Addition with Decimals
Subtraction with Decimals
Multiplication with Decimals
Fraction Arithmetic
Recognizing Primes
Long Division for Decimals
Square Root Simplification
Greatest Common Divisors
Least Common Multiples

Words Before Symbols: 
What is a Variable?
Introduction
Variation between Examples

Variation of Letters

A letter denotes a variable

Cases of Double Variation

Three Notions of a Variable

Constants, Parameters
& Variables

Talking about numbers
Dependent or Independent
Variable, a Matter of Choice

Complex number: starter lesson  

Solving Linear Equations:

A. Letters and Lengths

B. & C. Solving Linear Eq'ns
with stick diagrams.

(i) x + 20 = 29
(ii) 2x + 5 = 20
(iii) 3x + 10 = 32
(iv) 5a + 16 = 3a+ 24

(v)  (½)x + 8 = 24½
(vI)  (¾)a + 16 = (¼)a+ 24
(vii) (¾)q + 17 = 32
(viii) 13 =[2/3]x +7 twice
(x) Animated Examples
(i) Integral Coefficients (A)
(ii) Integral Coefficients (B)
(iii) Fractional Coefficients

(iv) With Parameters

Problem Solving with Linear
Equations in one or many
unknowns, and in essentially 
one unknown - Symbols before
words. 


C. Solving Linear Eq'ns 
without
Stick Diagrams

D. Problems in 
essentially one unknown

E: 2D Systems - Sub Methods.
F. Larger Systems



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