Convert recurrent formula with polynomial term / parameter to explicit formula So, I know how to convert to explicit formulas things like the Fibonacci sequence cause it only consists of $a_n$ like this:
$$a_{n} = a_{n-1} + a_{n-2}$$
However my problem is I've encountered a type of this problem I haven't been thought how to approach, and can't find the solution anywhere:
$$a_{n} = a_{n-1} + n+1$$
The sequence this is supposed to represent is: $0, 2, 5, 9, 14, 20, 27 ...$
The correct explicit formula that I don't know how to get to for this is:
$$a_n = \frac{n}{2}(n+3)$$

To solve this, I've tried converting it to:
$$r^n = r^{n-1} + n + 1$$
and treating n as the superscript of r... to end up with
$$r^2=r+2+1$$
..which is the standard procedure I've been taught, then solving for $r$ to get $r_1$ and $r_2$ and adding $\alpha$ and $\beta$ like I've been taught to get:
$$a_n = \alpha(r_1)^n + \beta(r_2)^n$$
.. then plugging in known $n$ and $a_n$ values to get a system of equations and then finally plug in the resulting $\alpha$ and $\beta$ to get the explicit formula, which however turned out to be complete gibberish. I'm sure I didn't mess up the system of equations or any step since I used automated equation solving to make sure.
That means the problem is me not knowing how to deal with that problem in the first place, I think. It seems to have a non-standard procedure to it.
 A: From the original recurrence relation
$$a_n=a_{n-1}+n+1$$
$$=(a_{n-2}+n)+n+1$$
$$=((a_{n-3}+(n-1))+n)+n+1$$
Continuing this pattern and knowing that $a_1=0$, we then have that
$$a_n=\sum_{r=2}^{n+1}r=\frac{1}{2}(n+1)(n+2)-1=\frac{1}{2}(n^2+3n+2)-1=\frac{1}{2}(n^2+3n)=\frac{n}{2}(n+3)$$
You can also approach this in the way you stated but we must first solve the associated homogeneous recurrence relation
$$a_n=a_{n-1}$$
and then find the particular solution
$$a_n=\lambda n+\mu$$
by using this definition for $a_n$ in the original recurrence relation.
A: If $g(x) = \sum_{n=0}^\infty a_n x^n$ is the generating function of your sequence (with $a_0 = 0$), you have
$$ g(x) = \sum_{n=1}^\infty (a_{n-1} + n + 1) x^n = x g(x) + \sum_{n=1}^\infty (n+1) x^n = x g(x) + \frac{x(2-x)}{(1-x)^2}$$
so that 
$$g(x) = \frac{x (2-x)}{(1-x)^3}$$
Convert this to partial fractions:
$$ g(x) = \frac{1}{(1-x)^3} - \frac{1}{1-x} $$
Note that $1/(1-x) = \sum_{n=0}^\infty x^n$ while 
$$ \frac{1}{(1-x)^3} = \frac{1}{2} \frac{d^2}{dx^2} \frac{1}{1-x} = \sum_{n=0}^\infty \frac{(n+1)(n+2)}{2} x^n $$
A: The formula which you give, $a_n=\frac{n}{2}(n+3)$ is not the correct formula for the sequence defined recursively by $a_1=0, a_{n+1}=a_n+n+1$.
This is a sequence with a constant second difference.
The sequence is
$$ 0, 2, 5, 9, 14, 20, 27,\cdots $$
The first difference is found by subtracting each term from the following term:
$$ 2, 3, 4, 5, 6, 7, \cdots $$
The second difference is a constant sequence consisting entirely of ones.
When the second difference is constant, $a_n$ is a second degree polynomial in $n$
$$a_n=rn^2+sn+t$$
We are given that
$$ a_{n+1}-a_n=n+1 $$
Therefore
\begin{eqnarray}
r(n+1)^2+s(n+1)+t-(rn^2+sn+t)&=&n+1\\
2rn+r+s&=&n+1
\end{eqnarray}
So
\begin{eqnarray}
2r&=&1\\
r+s&=&1
\end{eqnarray}
which gives $r=s=\frac{1}{2}$
Now we have that
\begin{eqnarray}
a_n&=&\frac{1}{2}n^2+\frac{1}{2}n+t\\
a_1&=&\frac{1}{2}+\frac{1}{2}+t=0\\
\end{eqnarray}
Therefore, $t=-1$
\begin{eqnarray}
a_n&=&\frac{1}{2}n^2+\frac{1}{2}n-1\\
&=&\frac{1}{2}(n^2+n-2)\\
&=&\frac{1}{2}(n-1)(n+2)
\end{eqnarray}
