How can we determine the sum of a series where each term is a product of two integers using method of differences? I have been trying to find out the sum of a series up to the $n^{th}$ term but failed.$$S_n=1 \cdot 3+2 \cdot 4+3 \cdot 5+4 \cdot 6+ \ldots + n(n+2)$$
my work:
\begin{align*}
S_n & =\frac{n(n+2)(n+4)-(n-2)n(n+2)}{6}\\
    & =\frac{n(n+2)(n+4)}{6}-\frac{n(n+2)(n-2)}{6}\\
    & =V_n - V_0
\end{align*}
since $V_0=0$
$$S_n =\frac{n(n+2)(n+4)}{6}$$
but later I found that the actual sum is $$S_n=\frac{n(n+1)(2n+7)}{6}$$
Where did I make a mistake? I tried to find the sum in another way and that is by taking the sum of $n^2+2n$ and obtained the actual sum but couldn't find where I made a mistake in my first approach.
I am stuck with another problem too and that one is
$$S_n=1 \cdot 2+2 \cdot 5+3 \cdot 8+ \ldots + n(3n-1)$$
In this one I am completely clueless what to do since $n$ and $3n-1$ doesn't seem to be following a pattern between them.

The simplest way to solve such questions is to use the known results for $\sum i, \sum i^2, ...$.
So, for your second problem you have to find $3\sum i^2-\sum i.$
In your comments you say that you wanted to use the method of differences. This is possible and is a good method for some problems but in your first question you would have needed to obtain  a function $F$ such that $F(n)-F(n-1)=n(n+2)$. That function $F$ would actually have been $\frac{n(n+1)(2n+7)}{6}$, a far from easy function to find!
You would then have the required form for the method of differences:
$$\frac{n(n+1)(2n+7)}{6}-\frac{(n-1)n(2n+5)}{6}=n(n+2).$$
