find nth term of a few sequences first $4,-1,-11,-26,-46$
I found a recurrence relation of $U_{n-1} - 5(n-1)$ but I don't know how to find an explicit nth term formula 
second $0,3,8,15,24$
It goes up by $+3,+5,+7,+9$ I'm not sure how to find nth term for this
 A: The sequence is $$-\left(\frac{5n(n + 1)}{2} - 4\right).$$ The OEIS is a terrific website, but, like any search engine, you occasionally have to finesse your search terms. Like Aleksandar showed, your sequence as you've presented it here is not in the OEIS. But try removing the initial $4$ and you get http://oeis.org/A166137 which is described as 5*n*(n+1)/2-4. The offset is 1,2. I don't know what the 2 means, but I do know the 1 means that the first $n$ is $n = 1$. If you try $n = 0$ in 5*n*(n+1)/2-4 you get $-4$.
A: The second one is $a_n=n^2-1$ for $n=1,2,3,...$
EDIT: my formula for the first was wrong; it's $a_n=4-5\cdot\frac{n(n+1)}{2}$ for $n=0,1,2,...$
A: Friend here is a website that you might find useful it is called OEIS or the Online Encyclopedia of Integer Sequences it is very useful.
Just type in your integer sequences and you are good to go. Here is the results of a few OEIS searches.
$4,-1,-11,-26,-46$
The sequence didn't appear to be indexed but it claimed to be:
$a_{n}=\frac{-5}{2}n^{2}+\frac{5}{2}n+4$
Where $a_{0}=4$, $a_{1}=-1$, $a_{2}=-11$, $a_{3}=-26$ and so on.
The next sequence was,
$0,3,8,15,24$
It can be multiple sequences here are a few sequences:
$a_{n}=n(n+2)$ 
$a_{n}=(n+1)^2-1$
If you want to know why it is so, plug in a value for $n$ and it will satisfy your sequence.
