Reasonably simple series question

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In summary, if you invest £1000 on the first day of each year and interest is paid at 5% on your balance at the end of each year, after 25 years you will have £10,525.
  • #1
henryc09
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Homework Statement


If you invest £1000 on the first day of each year and interest is paid at 5% on your balance at the end of each year how much money do you have after 25 years?


Homework Equations





The Attempt at a Solution


Been a while since I did series, and I was never very good with them. I'm guessing you have some sort of combination of an arithmetic and geometric series, just not really sure how to set it up. Any points in the right direction would be appreciated
 
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  • #2
Lets take a look at each year separately, I'll denote the nth year as Pn.

[itex]P_0=1000[/itex] : since you invested £1000 at the beginning of the year.

[itex]P_1=1000+1000(0.05)[/itex] : you already have 1000 in the bank, but now you get 5% interest on that money, so you need to add 0.05x of the investment you have to the bank.

[itex]P_2=P_1(0.05)+P_1[/itex] : now notice that we can factorize P2

[itex]P_2=P_1(1+0.05)=P_1(1.05)[/itex] : and in the same way we can factorize P1.

[itex]P_1=P_0(0.05)+P_0=P_0(1.05)[/itex]

So now, [itex]P_2=P_0(1.05)^2[/itex] and following the same pattern, [itex]P_3=P_0(1.05)^3[/itex] etc.

So after 25 years?
 
  • #3
I don't think this accounts for the fact that you add another £1000 at the beginning of each year though
 
  • #4
Look at the amount added each year separately.
The first years £1000 becomes 1000+ 1000(.05)+ 1000(.052)+ ...+ 1000(.0525) which is a geometric series.

The second years £1000 becomes 1000+ 1000(.05)+ 1000(.052)+ ...+ 1000(.0524) which is also a geometric series, etc.

The last years contribution will be just 1000+ 1000(.05).

Each of those can be calculsted by the formula:
[tex]\sum_{i=0}^n ar^i= a\frac{1- r^{n-1}}{1- r}[/tex]

For the first year's £1000, that is
[tex]1000\frac{1- .05^{24}}{1- .05}= \frac{1000}{.95}(1- .05^{24})= (1052.63)(1- .05^{24})[/tex]

For the second year's £, that is
[tex]1000\frac{1- .05^{23}}{1- .05}= \frac{1000}{.95}(1- .05^{23})= (1052.63)(1- .05^{23})[/tex]
etc.

Adding all those together, we have [tex](1052.63)(1- .05^{24}+ 1- .05^{23}+ ...+ 1- .05+ 1)= (1052.63)(25- (.05+ .05^2+ ...+ .05^{24})[/tex]
which gives a second geometric series.
 
  • #5
henryc09 said:
I don't think this accounts for the fact that you add another £1000 at the beginning of each year though
Oh sorry, I misread the question.
 
  • #6
I really don't think that works because the answer you get is far too small, I worked it out manually with a calculator and the answer is near £50,000
 
  • #7
Ok I'll do the question again, after reading all the information this time :redface:

[tex]P_0=1000[/tex]

[tex]P_1=P_0.I+1000=1000(1.05)+1000[/tex] : I denotes interest. You multiply by 1.05 rather than 0.05 because the interest of 50 is added to the 1000, so [itex]1000(0.05)+1000=1000(1+0.05)=1000(1.05)[/itex]

[tex]P_2=P_1.I+1000=\left(1000(1.05)+1000\right)(1.05)+1000[/tex]

Now, expand [itex]P_2[/itex] and see if you can spot the pattern. Try finding what [itex]P_3[/itex] is to be sure you're getting the pattern right. So what would be [itex]P_25[/itex]?

Now, do you know how to tackle the geometric series?
 
  • #8
I do believe, if I understand your question correctly, that the easiest method to solve this problem is by the use of this formula:

[tex]A = P \left(1 + \frac{r}{n}\right) ^{rt} \large [/tex]

Where:
P is the inital amount/investment
r is the interest rate as a decimal
n is the number of times the amount is compounded per year
t is the number of years
A is the amount after t years

Hopefully that helps.
 

Related to Reasonably simple series question

1. What is a "reasonably simple series question"?

A "reasonably simple series question" is a type of question that involves a sequence of numbers, letters, or other elements that follow a pattern or rule. These questions are commonly used in mathematics and logic puzzles.

2. How can I solve a reasonably simple series question?

To solve a reasonably simple series question, you need to carefully observe the given elements and identify the pattern or rule that governs the sequence. Once you understand the pattern, you can apply it to the next element in the series to find the missing term.

3. Is there a specific strategy or method for solving reasonably simple series questions?

There are several strategies and methods that can be used to solve reasonably simple series questions, such as finding the difference or ratio between consecutive terms, looking for repeating patterns, or using algebraic equations. It is important to practice and develop your own problem-solving approach.

4. Are there any tips for solving reasonably simple series questions more efficiently?

One tip for solving reasonably simple series questions is to start by looking at the first and last terms in the series. These can often provide clues about the pattern or rule. Additionally, it can be helpful to break down the series into smaller parts if it appears to have multiple patterns or rules.

5. Can reasonably simple series questions be used in real-world applications?

Yes, reasonably simple series questions can be used in real-world applications such as predicting stock market trends, analyzing data patterns in scientific research, and developing algorithms for technology. They can also help improve critical thinking and problem-solving skills in various fields.

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