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.

Firstly: You are confusing the sum of the sequence with its individual elements when you write
$$S_n =\frac{n(n+2)(n+4)-(n-2)n(n+2)}{6}$$
Let us write $S_n=u_1+\cdots+u_n$, where $u_k=k(k+2)$. So now we have
$$u_k =\frac{k(k+2)(k+4)-(k-2)k(k+2)}{6}$$
(which I think is what you meant).
Secondly: This is of the form $u_k=F(k)-F(k-2)$, with $F(k)=\frac16k(k+2)(k+4)$. But to make the method of differences work, you need to express $u_k$ as $F(k)-F(k-1)$, not as $F(k)-F(k-2)$.
But all is not lost! We can sum the odd-numbered elements and the even-numbered elements separately using your formula. We get:
$$u_1+u_3+\cdots+u_{2m-1}=F(2m-1)-F(1)+u_1=F(2m-1)+\frac12$$
$$u_2+u_4+\cdots+u_{2m}=F(2m)-F(2)+u_2=F(2m)$$
So if $n=2m$ is even, we get
$$S_n=u_1+\cdots u_n=(F(n-1)+\frac12)+(F(n))$$
And if $n=2m+1$ is odd, we get
$$S_n=u_1+\cdots u_n=(F(n)+\frac12)+(F(n-1))$$
In both cases,
$$S_n=F(n)+F(n-1)+\frac12$$
If you do the algebra, you will see that this is equal to
$$\frac{n(n+1)(2n+7)}{6}$$
as required.
