Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$ The question
Sum the first $n$ terms of the series:
$$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$
This was asked under the heading using method of difference and the answer given was
$$ S_n = \frac{1}{10}n(n+1)(n+2)(n+3)(2n+3). $$
My approach
First, I get $$ U_n=n(n+2)(n+1)^2. $$
Then I tried to make $U_n = V_n - V_{n-1}$
in order to get $S_n = V_n - V_0$.
But I really don't know how can I figure this out.

\begin{align*}
\sum_{k=1}^n k(k+2)(k+1)^2&=\sum_{k=1}^n (k+3)(k+2)(k+1)k-2\sum_{k=1}^n (k+2)(k+1)k\\
&=24 \sum_{k=1}^n \binom{k+3}{4}-12 \sum_{k=1}^n \binom{k+2}{3}\\
&=24\binom{n+4}{5}-12\binom{n+3}{4}
\end{align*}
