Geometric sequence, find the best interest option over a year

This is what you are trying to find.In summary, the Bank of Utopia offers a 100% interest rate per annum with various options for adding the interest. A man invests $1000 and considers different options, including adding interest annually, half-yearly, or as often as required. Investigating the maximum possible amount in the man's account after one year, it is found that the amount increases to infinity as the number of credits per year increases. However, the differences between the amounts tend towards zero. The problem involves finding the maximum possible amount by using a geometric series and calculating its sum to infinity. Further research suggests the use of Napier's constant, also known as the base of natural logarithms, to solve
  • #1
thekopite
5
0

Homework Statement



The Bank of Utopia offers an interest rate of 100% per annum with various options as to how the interest may be added. A man invests $1000 and considers the following options.
Option A - Interest added annually at the end of the year.
Option B - Interest of 50% credited at the end of each half-year.
Option C, D, E, ... The Bank is willing to add interest as often as required, subject to (interest rate) x (number of credits per year) = 100
Investigate to find the maximum possible amount in the man's account after one year.

Homework Equations



The Attempt at a Solution



So I took 1000(1 + (1/x)^x as the amount in the man's account by the end of the year, where x is the number of credits per year. I'm fairly sure this amount increases to infinity as x increases, but the differences between the amounts as x (x remaining an integer) increases must tend toward zero (considering the question). Since this is a section on geometric series I'm wondering if I'm supposed to salvage a geometric series out of this and calculate it's sum to infinity, but I have no idea which series to look for. Any suggestions would be appreciated, cheers.
 
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  • #2
thekopite said:
So I took 1000(1 + (1/x)^x as the amount in the man's account by the end of the year, where x is the number of credits per year. I'm fairly sure this amount increases to infinity as x increases...
Actually, the amount doesn't increase to infinity. Try putting in larger and larger values of x into a graphing calculator or spreadsheet and see what happens.
 
  • #3
thekopite said:

Homework Statement



The Bank of Utopia offers an interest rate of 100% per annum with various options as to how the interest may be added. A man invests $1000 and considers the following options.
Option A - Interest added annually at the end of the year.
Option B - Interest of 50% credited at the end of each half-year.
Option C, D, E, ... The Bank is willing to add interest as often as required, subject to (interest rate) x (number of credits per year) = 100
Investig!te to find the maximum possible amount in the man's account after one year.

Homework Equations



The Attempt at a Solution



So I took 1000(1 + (1/x)^x as the amount in the man's account by the end of the year, where x is the number of credits per year. I'm fairly sure this amount increases to infinity as x increases, but the differences between the amounts as x (x remaining an integer) increases must tend toward zero (considering the question). Since this is a section on geometric series I'm wondering if I'm supposed to salvage akgeometric series out of this and calculate it's sum to infinity, but I have no idea which series to look for. Any suggestions would be appreciated, cheers.

Hmm...sounds like they want you to discover the letter 'e'. :biggrin:

Look up Napier's constant, also known as the base of natural logarithms.
 

Related to Geometric sequence, find the best interest option over a year

1. What is a geometric sequence?

A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant value, called the common ratio. The general form of a geometric sequence is a, ar, ar^2, ar^3, ... where a is the first term and r is the common ratio.

2. How do I find the common ratio in a geometric sequence?

To find the common ratio in a geometric sequence, divide any term by the previous term. The resulting value will be the common ratio.

3. What is the best interest option for a geometric sequence over a year?

The best interest option for a geometric sequence over a year depends on the specific terms of the sequence and the available interest rates. To find the best option, you can compare the interest rates offered by different financial institutions or use an online calculator to determine the total interest earned for each option.

4. How can I calculate the total interest earned in a year for a geometric sequence?

To calculate the total interest earned in a year for a geometric sequence, you can use the formula total interest = initial amount x (1 + interest rate)^number of terms - initial amount. Make sure to use the correct interest rate and number of terms in the formula.

5. Can a geometric sequence have a negative common ratio?

Yes, a geometric sequence can have a negative common ratio. This means that each term in the sequence will be a negative value. However, the sequence will still follow the same pattern of multiplying each term by the common ratio to get the next term.

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