## Task 21/76

Calculate the product: $\prod_{i=1}^{n} \sum_{j=1}^{i} j^{2}$.
Certainly, here is the translation of the provided text into English, preserving the original formatting:

```
It is well known that $\sum_{j=1}^{i} j^{2}=\frac{1}{6} i(i+1)(2 i+1)$. Therefore,

$$
\begin{aligned}
\prod_{i=1}^{n} \sum_{j=1}^{i} j^{2}= & \prod_{i=1}^{n} \frac{1}{6} i(i+1)(2 i+1)=\prod_{i=1}^{n}(i+1) \cdot \prod_{i=1}^{n} \frac{1}{12} \cdot \prod_{i=1}^{n} 2 i(2 i+1)= \\
& =(n+1)!\cdot \frac{1}{12^{n}} \prod_{i=1}^{2 n+1} i=\frac{1}{12^{n}}(n+1)!(2 n+1)!
\end{aligned}
$$
```