Is there another way to solve this problem?

[Darya]

The problem: At the beginning of each of the four years, we deposit € 500 in savings account. At the end of each year, the bank will credit us with interest of 4,5%. Calculate what amount will be on your account after accruing interest at the end of the fourth year if we don't pay any additional fees or taxes.

So since every year we're gonna add 500 more euros to our account, the solution is 500*1,045+(500+500*1,045)*1,045 and so on till the 4th year... I'm wondering whether there's a faster and more rational way to calculate it. Thanks

 

[Steven G]

Did you try factoring?

 

[Mr. Bland]

I actually ran into something very similar the other day and it's neither intuitive nor succinct. It's a problem of compound interest with recurring payments. As such, I have reservations about the general solution being involved with this assignment. Is the problem asking for only the answer after four years?

Each year, the balance in the account is modified like this:

[MATH]balance = (previous + payment) * (1 + interest) = (previous + 500) * 1.045[/MATH]​

Mr. Spreadsheet tells us the balances at the end of each year will be as follows:

• 522.50
• 1068.51
• 1639.10
• 2235.35

Calculating this over an arbitrary number of years involves two steps: finding the principal given the interest beginning with the second year, then adding the amount of the recurring payments with interest. Rolled together, the formula looks like this:

[MATH]total = payment\frac{\left((1 + interest)^{years} - 1\right)(1 + interest)}{interest}[/MATH]​

I'm still not fully clear on the relationship between the reciprocal of the interest rate and the amount contributed by the recurring payment, so I'm sort of a delinquent for bringing it up. I should really brush up on how to actually solve a recursive expression like this...

 

[JeffM]

There is a formula used by financial professionals, namely

[MATH]f = \text {future value of annuity due at the end of the last period.}[/MATH]
[MATH]a = \text {amount of payment due at the start of each period.}[/MATH]
[MATH]n = \text {number of periods.}[/MATH]
[MATH]i = \text {interest rate PER PERIOD expressed as a decimal.}[/MATH]
Note that if there are multiple periods per year, i is not the annual interest rate.

[MATH]f = a(1 + i) * \dfrac{(1 + i)^n - 1}{i}, \text { provided that } i > 0.[/MATH]
If i = 0, then we have a simple multiplication problem because [MATH]f = a * n.[/MATH]
Let's put that into your problem, where a = 500, i = 0.045, and n = 4. (HERE, the periods are one year in length so in this special case i is equal to the annual interest rate.)

[MATH]500 * 1.045 * \dfrac{(1.045)^4 - 1}{0.045} \approx \dfrac{522.5}{0.045} * (1.1925 - 1) \approx 2235.35486 \approx 2235.35.[/MATH]
Let's check.

At the end of first year, the value is [MATH]500 * 1.045 = 522.50000.[/MATH]
At the end of second year, the value is [MATH](500 + 522.50000) * 1.045 = 1068.51250.[/MATH]
Third year, [MATH](500 + 1068.51250) * 1.045 \approx 1639.09556.[/MATH]
Final year year, [MATH](500 + 1639.09556) * 1.045 \approx 2235.35486 \approx 2235.35.[/MATH]
The formula may look difficult to use, but suppose you had monthly payments for 10 years. In any case, if you have to do lots of problems like this, you simply program your computer to use the formula.

The derivation of the formula is a bit more complex than I suspect you to be ready for.

 

[MarkFL]

At the end of year \(n\) where \(A_n\) is the amount in the account and \(D\) is a fixed annual deposit, and \(r\) is the annually compounded interest rate, we have:

[MATH]A_{n}=(1+r)\left(A_{n-1}+D\right)[/MATH]
This can be arranged as the following inhomogeneous linear difference equation:

[MATH]A_{n}-(1+r)A_{n-1}=(1+r)D[/MATH]
The homogeneous solution is:

[MATH]h_n=c_1(1+r)^n[/MATH]
The particular solution is:

[MATH]p_n=A[/MATH]
Substituting this into our difference equation, there results:

[MATH]A-(1+r)A=(1+r)D[/MATH]
[MATH]A=-\frac{1+r}{r}D[/MATH]
And so the general solution to the difference equation is:

[MATH]A_n=h_n+p_n=c_1(1+r)^n-\frac{1+r}{r}D[/MATH]
To determine the value of the parameter, we can use the initial value:

[MATH]A_0=c_1-\frac{1+r}{r}D=0\implies c_1=\frac{1+r}{r}D[/MATH]
And so our solution, or closed form, becomes:

[MATH]A_n=\left(\frac{1+r}{r}D\right)(1+r)^n-\frac{1+r}{r}D[/MATH]
[MATH]A_n=\left(\frac{1+r}{r}D\right)\left((1+r)^n-1\right)[/MATH]
Using the values given in the problem, we find:

[MATH]A_4=\left(\frac{1+0.045}{0.045}D\right)\left((1+0.045)^4-1\right)\approx2235.35[/MATH]