3. Find the sum: $S_{n}=1^{2}+2^{2}+3^{2}+\cdots+n^{2}$.
3. Solution: Add $0=k-k$ after each term, i.e.,
$$
\begin{array}{l} 
\quad k^{2}=k^{2}+k-k=k(k+1)-k=2 C_{k-1}^{2}-k, k=1,2,3 \cdots n \\
S_{n}= \sum_{k=1}^{n} k^{2}=2 \sum_{k=1}^{n} C_{k+1}^{2}-\sum_{k=1}^{n} k \\
= 2\left(C_{2}^{2}+C_{3}^{2}+\cdots+C_{n+1}^{2}\right)-(1+2+3+\cdots+n) \\
= 2 C_{n+2}^{3}-\frac{1}{2} n(n+1)=\frac{1}{6} n(n+1)(2 n+1) .
\end{array}
$$