Example 5. Given $a_{1}=3, a_{n+1}=a_{n}+2 n+3$ $(n \geqslant 1)$, find the general term $a_{n}$ of the sequence.
Solution 1
$$
\begin{array}{l}
\because a_{1}=3, \quad a_{n+1}=a_{n}+2 n+3, \\
\therefore \mathrm{a}_{2}=8=2 \times(2+2) \text {, } \\
\mathrm{a}_{3}=15=3 \times(3+2) \text {, } \\
a_{4}=24=4 \times(4+2) \text {, } \\
\text {... } \\
a_{n}=\left\{\begin{array}{ll}
3, & (n=1) \\
n & (n+2) .(n>1)
\end{array}\right. \\
\end{array}
$$

Proof omitted.
Solution 2 From $a_{1}=3, a_{n+1}=a_{n}+2 n+3$,
$$
\begin{array}{l}
(n \geqslant 1) \\
a_{1}=3 \\
a_{2}=a_{1}+2 \times 1+3 \\
a_{3}=a_{2}+2 \times 2+3 \\
a_{4}=a_{3}+2 \times 3+3 \\
\cdots \cdots \cdot \\
+a_{n}= a_{n-1}+2(n-1)+3 \\
S_{n}= S_{n-1}+2[1+2+3+\cdots+ \\
(n-1)]+3 n .
\end{array}
$$

Therefore,
$$
\begin{aligned}
a_{n} & =S_{n}-S_{n-1} \\
& =2[1+2+3+\cdots \\
& +(n-1)]+3 n \\
& =n^{2}+2 n=n(n+2) .
\end{aligned}
$$