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

Notice, $$T_n=T_{n-1}+3$$
setting $n=1$, $$T_1=T_0+3=1+3=4$$
$n=2$, $$T_2=T_1+3=4+3=7$$
$n=3$, $$T_3=T_2+3=7+3=10$$
$n=4$, $$T_4=T_3+3=10+3=13$$
$$........................$$
$$........................$$
$$T_n=T_{n-1}+3$$
thus, one should observe that an A.P. is obtained with common difference $d=3$ & the first term $a=4$ hence $n$th term of A.P. $$\color{red}{T_n}=a+(n-1)d$$$$=4+(n-1)3=\color{red}{3n+1}$$
