Plugging number back into recurrence relation I have this problem that I already solved the recurrence for:
$$T_{n} = T_{n-1} + 3, T_{0} = 1$$
I worked it out to $T_{n-4} + 4[(n-3)+(n-2)+(n-1)+n]$ (where I stopped because I saw the pattern), but I can't figure out how to actually plug in the 1 from the $T_{0} = 1$. 
If I remember correctly, the $T_{n-4} + 4[(n-3)+(n-2)+(n-1)+n]$ works out to be $T_{n-k} + 4[(n-3)+(n-2)+(n-1)+n]$.

Here's a simple way to see this :
$$T_n=T_{n-1}+3\implies T_{n-1}=T_{n-2}+3\implies\cdots\implies T_2=T_1+3\implies T_1=T_0+3$$
So, using this chain of implications (summing them up), you can write,
$$T_n+(T_{n-1}+T_{n-2}\cdots+T_1)=(T_{n-1}+T_{n-2}+\cdots+T_1)+T_0+\left(\sum_{i=1}^n 3\right)$$
Cancelling stuff from both sides,
$$T_n=\left(\sum_{i=1}^n 3\right)+T_0\implies T_n=3n+1~\forall~n\geq 0$$
