15. $\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\}$ are three sequences satisfying $a_{1}=1, b_{1}=3, c_{1}=2$ and
$$
\left\{\begin{array}{l}
a_{i+1}=a_{i}+c_{i}-b_{i}+2 \\
b_{i+1}=\frac{1}{2}\left(3 c_{i}-a_{i}+5\right) \\
c_{i+1}=2 a_{i}+2 b_{i}-3 \\
(i=1,2, \cdots)
\end{array}\right.
$$

Find $\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\}$.
15. (By observing several items, we can guess)
$a_{i}=2^{i-1}, b_{i}=2^{i}+1, c_{i}=3 \cdot 2^{i-1}-1$. Use mathematical induction. Assume it holds for $i=n$, i.e., $a_{n}=2^{n-1}, b_{n}=2^{n}+1, c_{n}=3 \cdot 2^{n-1}-1$.
Then $a_{n+1}=a_{n}+c_{n}-b_{n}+2$
$$
\begin{aligned}
& =2^{n-1}+3 \cdot 2^{n-1}-1-2^{n}+2=2^{n} \\
b_{n+1}= & \frac{1}{2}\left(3 c_{n}-a_{n}+5\right) \\
= & \frac{1}{2}\left(9 \cdot 2^{n-1}-3-2^{n-1}+5\right)=2^{n+1}+1 \\
c_{n+1}= & 2 a_{n}+2 b_{n}-3 \\
= & 2^{n}+2^{n+1}+2-3=3 \cdot 2^{n}-1
\end{aligned}
$$

Therefore, the induction holds, so the general terms of $\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\}$ are $a_{n}=2^{n-1}, b_{n}=2^{n}+1$,
$$
c_{n}=3 \cdot 2^{n-1}-1 \text {. }
$$