A rather simple example is this one:
1 4 9 16 25 36 49 ....
3 5 7 9 11 16 ....
2 2 2 2 2 ....
We can clearly see it's:
$$ (i+1)^2 $$
but my question is how to do find the formula in general? I'm struggling with this one:
2 3 5 8 12 17 23 ....
1 2 3 4 5 6 ....
1 1 1 1 1 ....
at the moment but maybe learning a general approach first will help.
On
By the method you are doing, if you keep calculating successive differences and you encounter an A.P like above. Then you know that the general term of an A.P is of the form $a_{n}=an+b$, then the sequence above will have a quadratic general term of the form $a_{n}=an^2+bn+c$ and so on.
Like for your second sequence, the first differences are in A.P, therefore the sequence above will have a quadratic general term.
On
Let $f$ be the second sequence we are interested in with terms indexed from zero. Define $(\Delta f)(n)=f(n+1)-f(n)$ for $n\geq 0$ which is sometimes called the first difference. Note that the operator $\Delta $ produces another sequence which in the diagram is the sequence below the desired sequence.
We may apply $\Delta$ again to this sequence to produce the sequence $\Delta^2 f$ which is sometimes called the second difference) and this sequence is written below the first difference in the diagram above. In general we can continue to compute $\Delta ^m f$ for $m\geq1$.
Before we get to the question also note that if $\Delta f=0$ iff $f(n)=c$ for some $c$. One direction is obvious. For the other note that $\Delta f=0$ implies that $f(n+1)=f(n)$ for all $n\geq 0$. In particular $f(n+k)=f(n)$ for $k\geq 1$, by repeatedly applying the previous equality, whence $f(n)=f(0)$ for all $n$. As a corollary $\Delta f= \Delta g$ iff $f(n)=g(n)+k$ for some $k$.
Now onto the problem. Note that $$ (\Delta^2 f)(n)=1\iff(\Delta f )(n)=n+c$$ for some constant $c$. But $(\Delta f)(0)=1$ whence $c=1$. Now $$ (\Delta f )(n)=n+1\iff f(n)=\frac{1}{2}n(n-1)+n+k\tag{1} $$ for some $k$. But $f(0)=2$ whence $k=2$. Thus $$ f(n)=\frac{1}{2}n(n-1)+n+2;\quad (n\geq 0). $$ In (1) we used the fact that $$ \Delta n^\underline{m}=m\times n^\underline{m-1} ;\quad (m\geq1) $$ where $n^\underline{m}=n(n-1)\dotsb(n-m+1)$ is the falling factorial of length $m$.
The process we used is entirely analogous to finding antiderivatives.
You want to find a general formula for $(c_n)=2,3,5,8,12,17,23,30,...$?
Observe that $c_1=2$ and $c_n=(n-1)+c_{n-1}$ for $n\geq 2$ and thus $$c_n=c_1+\sum_{i=1}^{n-1}i=2+\frac{(n-1)n}{2}$$