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

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MarkFL

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Feb 24, 2012
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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.
 
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MarkFL

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Feb 24, 2012
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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:

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