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.
 A: 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:
$$\frac{n(n+1)(2n+7)}{6}-\frac{(n-1)n(2n+5)}{6}=n(n+2).$$
A: 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.
A: $S_n$ will be a cubic function of $n$, because there are $n$ terms of $2nd$ degree (quadratic). Then, we can write the formula for $S_n$ as $S_n = An^3 + Bn^2 + Cn + D$.
When $n=0$, $S_n = 0\cdot2 = 0 \Rightarrow D = 0$
When we plug in $n=1,2,3$, we get this system of equations:
\begin{align}
\tag{1}
A+B+C = 3
\newline
\tag{2}
8A+4B+2C = 11
\newline
\tag{3}
27A+9B+3C = 11
\end{align}
Solving this system gives $A=\frac{1}{3}, B = \frac{3}{2}, C = \frac{7}{6}$. Plugging these in to $S_n$, we get:
\begin{align}
S_n = \frac{2n^3+9n^2+7n}{6} = \frac{n(n+1)(2n+7)}{6}
\end{align}
