The seventh question: For an integer $n \geq 2$, non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfy $\sum_{i=1}^{n} a_{i}=4$. Try to find the maximum possible value of $2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+ \ldots +a_{1} a_{2} \ldots a_{n}$.
The seventh question:
Answer: When $a_{1}=3, a_{2}=1, a_{3}=a_{4}=\ldots=a_{n}=0$, $2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\ldots+a_{1} a_{2} \ldots a_{n}=9$.
Below is the proof: $2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\ldots+a_{1} a_{2} \ldots a_{n} \leq 9$.
(1) When $n=2$, $2 a_{1}+a_{1} a_{2}=a_{1}\left(2+a_{2}\right) \leq \frac{\left[a_{1}+\left(2+a_{2}\right)\right]^{2}}{4}=9$.
(2) When $n=3$, $2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}=2 a_{1}+a_{1} a_{2}\left(1+a_{3}\right) \leq 2 a_{1}+a_{1}\left[\frac{a_{2}+\left(1+a_{3}\right)}{2}\right]^{2}=$ $2 a_{1}+\frac{a_{1}\left(5-a_{1}\right)^{2}}{4}=9-\frac{\left(4-a_{1}\right)\left(a_{1}-3\right)^{2}}{4} \leq 9$.
(3) When $n=4$, we consider two cases.
$$
\begin{array}{l}
\text { (1) If } a_{3} \geq 1, \text { then } a_{1} \leq 3, \text { so } \\
2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+a_{1} a_{2} a_{3} a_{4}=2 a_{1}+a_{1} a_{2}\left(1+a_{3}+a_{3} a_{4}\right) \\
\leq 2 a_{1}+a_{1} a_{2}\left(1+a_{3}\right)\left(1+a_{4}\right) \\
\leq 2 a_{1}+a_{1}\left[\frac{a_{2}+\left(1+a_{3}\right)+\left(1+a_{4}\right)}{3}\right]^{3} \\
=2 a_{1}+\frac{a_{1}\left(6-a_{1}\right)^{3}}{27}=9-\frac{\left(a_{1}-3\right)^{3}\left(a_{1}-9\right)}{27} \leq 9
\end{array}
$$
(2) If $\mathrm{a}_{3} \leq 1$, then
$$
\begin{array}{l}
2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+a_{1} a_{2} a_{3} a_{4}=2 a_{1}+a_{1} a_{2}\left(1+a_{3}+a_{3} a_{4}\right) \\
\leq 2 a_{1}+a_{1} a_{2}\left(1+a_{3}+a_{4}\right) \\
\leq 2 a_{1}+a_{1}\left[\frac{a_{2}+\left(1+a_{3}+a_{4}\right)}{2}\right]^{2} \\
=2 a_{1}+\frac{a_{1}\left(5-a_{1}\right)^{2}}{4}=9-\frac{\left(4-a_{1}\right)\left(a_{1}-3\right)^{2}}{4} \leq 9
\end{array}
$$
(4) Assume the conclusion holds for $n-1(n \geq 5)$. When $n$, according to the AM-GM inequality, we have $a_{1} a_{2} \ldots a_{n-1} \leq$ $\left(\frac{a_{1}+a_{2}+\ldots+a_{n-1}}{n-1}\right)^{n-1} \leq\left(\frac{4}{n-1}\right)^{n-1} \leq 1<2$. Therefore, by the induction hypothesis, we have:
$$
\begin{array}{l}
9 \geq 2\left(a_{1}+a_{n}\right)+\left(a_{1}+a_{n}\right) a_{2}+\left(a_{1}+a_{n}\right) a_{2} a_{3}+\ldots+\left(a_{1}+a_{n}\right) a_{2} \ldots a_{n-1} \\
\geq 2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\ldots+a_{1} a_{2} \ldots a_{n-1}+2 a_{n} \\
\geq 2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\ldots+a_{1} a_{2} \ldots a_{n-1}+a_{1} a_{2} \ldots a_{n}
\end{array}
$$

In conclusion, the maximum value of $2 a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\ldots+a_{1} a_{2} \ldots a_{n}$ is 9.