15. (12 points) A person takes a loan of $s$ yuan from the bank to buy a house, with an annual interest rate of $p$. The person returns the principal and interest to the bank every two years, and pays off the loan in ten years (requiring the payment amount to the bank to be the same each time). What is the total amount paid over ten years?
Three, 15. Let the principal owed to the bank at the $i$-th payment be $s_{i}$ yuan, and the principal returned in the payment be $a_{i}$ yuan, with the interest being
$$
s_{i}\left[(1+p)^{2}-1\right] \text { (yuan). }
$$

According to the problem, the payment amount to the bank each time is the same, so we have
$$
\begin{array}{l}
s_{i}\left[(1+p)^{2}-1\right]+a_{i} \\
=\left(s_{i}-a_{i}\right)\left[(1+p)^{2}-1\right]+a_{i+1} .
\end{array}
$$

Rearranging gives $a_{i+1}=(1+p)^{2} a_{i}$.
The total principal repaid over five payments is $s$, so we have
$$
s=\frac{a_{1}\left[(1+p)^{10}-1\right]}{(1+p)^{2}-1},
$$

which means $a_{1}=\frac{s\left[(1+p)^{2}-1\right]}{(1+p)^{10}-1}$.
The amount of the first payment is
$$
\begin{array}{l}
\frac{s\left[(1+p)^{2}-1\right]}{(1+p)^{10}-1}+s\left[(1+p)^{2}-1\right] \\
=\frac{s\left[(1+p)^{2}-1\right](1+p)^{10}}{(1+p)^{10}-1} \text { (yuan). }
\end{array}
$$

Thus, the total amount paid over ten years is
$$
\frac{5 s\left[(1+p)^{2}-1\right](1+p)^{10}}{(1+p)^{10}-1} \text { (yuan). }
$$