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 $T_{n-k} + 4[(n-3)+(n-2)+(n-1)+n]$.

"Solving the recurrence" typically refers to obtaining an explicit formula for the $n$-th term of the recurrence. In this case, you should be able to show that $T_n=3n+1$. The "$+1$" arises from the $T_0=1$ condition.
