Question 15
Demonstrate that for all natural numbers $n\gt 1$ holds $$
1 \cdot \sqrt{\left(\begin{array}{l}
n \\
1
\end{array}\right)}+2 \cdot \sqrt{\left(\begin{array}{l}
n \\
2
\end{array}\right)}+\cdots+n \cdot \sqrt{\left(\begin{array}{l}
n \\
n
\end{array}\right)}\lt \sqrt{2^{n-1} n^{3}}
$$
1 \cdot \sqrt{\left(\begin{array}{l}
n \\
1
\end{array}\right)}+2 \cdot \sqrt{\left(\begin{array}{l}
n \\
2
\end{array}\right)}+\cdots+n \cdot \sqrt{\left(\begin{array}{l}
n \\
n
\end{array}\right)}\lt \sqrt{2^{n-1} n^{3}}
$$