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.
Best Answer
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.