10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$
10.6 .

It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$.