# Thread: Difficult interest rate problem

1. ## Difficult interest rate problem

I am not completely sure this involves calculus but it has been presented in our calculus class and I have tried looking at this from all different ways.

On June 9, 2007 you receive $700 as a graduation present from your wealthy Aunt Thrifty. She will give you$700 more each year that you leave it invested. You leave it invested until June 9, 2054 at the annual interest rate of 3.25%, compounded yearly. How much money is this?

I know you can do this year by year, but I need a formula. I can't figure out how to include the additional $700 she receives each year and the interest for it. 2. Originally Posted by cubs3205 I am not completely sure this involves calculus but it has been presented in our calculus class and I have tried looking at this from all different ways. On June 9, 2007 you receive$700 as a graduation present from your wealthy Aunt Thrifty. She will give you $700 more each year that you leave it invested. You leave it invested until June 9, 2054 at the annual interest rate of 3.25%, compounded yearly. How much money is this? I know you can do this year by year, but I need a formula. I can't figure out how to include the additional$700 she receives each year and the interest for it.
First year: 700 at 3.25% = 1.0325(700)=\$722.75
Second year: (700 at 3.25% + 700) at 3.25% = (1.0325)^2(700) + 1.0325(700)
Third year: [(700 at 3.25% + 700) at 3.25% + 700] at 3.25% = (1.0325)^3(700) + (1.0325)^2(700) + 1.0325(700)

N $_{th}$ year: $\sum^n_{k=1} 700(1.0325)^k$

From this we can do the total sum for 47 years as:

$\sum^{47}_{k=1} 700(1.0325)^k=\frac{700(1-1.0325^{47})}{1-1.0325}$

3. In the above response the reasoning is correct.
However, the last seems to be a bit off. See the below: r=1.0325

4. I will derive the answer (to see if what I wrote was mistaken).

Let

$S_{47}=\sum^{47}_{k=1} 700(1.0325)^k$

Then

$S_{47}=700(1.0325) + 700(1.0325)^2 +...+700(1.0325)^{47}$

$-1.0325S_{47}=-700(1.0325)^2-700(1.0325)^3-...-700(1.0325)^{47}-700(1.0325)^{48}$

Adding these two equations, we get:

$S_{47}-1.0325S_{47}=700(1.0325)-700(1.0325)^{48}=700\cdot1.0325(1-1.0325^{47})$

And so:

$S_{47}=(700\cdot 1.0325)\frac{1-1.0325^{47}}{1-1.0325}$

Thank you, Plato, for checking my work. I went off of memory (and apparently my memory is faulty). Next time I do a problem like this, I will just derive the formula so as to be less likely to make mistakes.

5. wow thanks. It works out well then.