4. Find the value of $\sqrt{14^{3}+15^{3}+16^{3}+\cdots+24^{3}+25^{3}}$
(A) 104
(B) 224
(C) 312
(D) 336
(E) 676
4. Answer: (C)
First, we note that $4 n^{3}=[n(n+1)]^{2}-[n(n-1)]^{2}$. Thus
$$
n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}-\left(\frac{n(n-1)}{2}\right)^{2} .
$$

Therefore
$$
\begin{aligned}
& 14^{3}+15^{3}+\ldots+24^{3}+25^{3} \\
= & \frac{14^{2} \times 15^{2}}{4}-\frac{13^{2} \times 14^{2}}{4}+ \\
& \frac{15^{2} \times 16^{2}}{4}-\frac{14^{2} \times 15^{2}}{4}+\ldots .+ \\
& \frac{24^{2} \times 25^{2}}{4}-\frac{23^{2} \times 24^{2}}{4}+ \\
& \frac{25^{2} \times 26^{2}}{4}-\frac{24^{2} \times 25^{2}}{4} \\
= & \frac{25^{2} \times 26^{2}}{4}-\frac{13^{2} \times 14^{2}}{4}=(25 \times 13+13 \times 7)(25 \times 13-13 \times 7) \\
= & (32 \times 13)(18 \times 13)=9 \times 64 \times 13^{2} .
\end{aligned}
$$

Thus $\sqrt{14^{3}+15^{3}+16^{3}+\ldots+24^{3}+25^{3}}=3 \times 8 \times 13=312$.