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:
$$F_{n+1}=(1+i)F_{n}+D$$ where $$F_0=D$$
where $i$ is the APR and $D$ is the annual deposit.
Now, we see the homogeneous solution is:
$$h_n=c_1(1+i)^n$$
and we seek a particular solution of the form:
$$p_n=A$$
Substituting the particular solution into the recurrence, we find:
$$A-(1+i)A=D\,\therefore\,A=-\frac{D}{i}$$
And so we have, by superposition:
$$F_{n}=h_n+p_n=c_1(1+i)^n-\frac{D}{i}$$
Now, using the initial value, we may determine the parameter $c_1$:
$$F_{0}=c_1-\frac{D}{i}=D\,\therefore\,c_1=\frac{D}{i}(1+i)$$
and so we have:
$$F_{n}=\frac{D}{i}\left((1+i)^{n+1}-1 \right)$$
To determine the amount $I_{n}$ of this that is interest, we must subtract the $n+1$ deposits that have been made:
$$I_{n}=F_{n}-(n+1)D=\frac{D}{i}\left((1+i)^{n+1}-1 \right)-(n+1)D$$
$$I_{n}=\frac{D}{i}\left((1+i)^{n+1}-(1+i(n+1)) \right)$$
Now, plugging in the data we are given for the problem:
$$D=3000,\,i=0.09,\,n=27$$
We find:
$$I_{27}=\frac{3000}{0.09}\left((1.09)^{28}-(1+0.09(28)) \right)\approx254904.65$$
