# Al's question at Yahoo! Answers regarding determining the accumulated interest

#### MarkFL

Staff member
Here is the question:

How to determine the accumulated value of interest?

Hey!

I have a maths problem that goes: "suppose you join a superannuation fund by investing \$3000 at 9% p.a. compound interest. The same amount is invested at the beginning of each subsequent year until you retire 27 years later. Determine the accumulated value of interest" I know the compound interest formula, however I have no idea how to account for the extra \$3000 that is added every year (on top of the 9% interest rate).

Thanks! I would be very grateful for any help
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello Al,

Let's let $F_n$ represent the value of the fund at the beginning of year $n$, where the initial year is year 0. We may model this situation with the recursion:

$$\displaystyle F_{n+1}=(1+i)F_{n}+D$$ where $$\displaystyle F_0=D$$

where $i$ is the APR and $D$ is the annual deposit.

Now, we see the homogeneous solution is:

$$\displaystyle h_n=c_1(1+i)^n$$

and we seek a particular solution of the form:

$$\displaystyle p_n=A$$

Substituting the particular solution into the recurrence, we find:

$$\displaystyle A-(1+i)A=D\,\therefore\,A=-\frac{D}{i}$$

And so we have, by superposition:

$$\displaystyle F_{n}=h_n+p_n=c_1(1+i)^n-\frac{D}{i}$$

Now, using the initial value, we may determine the parameter $c_1$:

$$\displaystyle F_{0}=c_1-\frac{D}{i}=D\,\therefore\,c_1=\frac{D}{i}(1+i)$$

and so we have:

$$\displaystyle F_{n}=\frac{D}{i}\left((1+i)^{n+1}-1 \right)$$

To determine the amount $I_{n}$ of this that is interest, we must subtract the $n+1$ deposits that have been made:

$$\displaystyle I_{n}=F_{n}-(n+1)D=\frac{D}{i}\left((1+i)^{n+1}-1 \right)-(n+1)D$$

$$\displaystyle I_{n}=\frac{D}{i}\left((1+i)^{n+1}-(1+i(n+1)) \right)$$

Now, plugging in the data we are given for the problem:

$$\displaystyle D=3000,\,i=0.09,\,n=27$$

We find:

$$\displaystyle I_{27}=\frac{3000}{0.09}\left((1.09)^{28}-(1+0.09(28)) \right)\approx254904.65$$