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 $T_{n-k} + 4[(n-3)+(n-2)+(n-1)+n]$.
 A: 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. $$\color{red}{T_n}=a+(n-1)d$$$$=4+(n-1)3=\color{red}{3n+1}$$
A: 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,
$$T_n=\left(\sum_{i=1}^n 3\right)+T_0\implies T_n=3n+1~\forall~n\geq 0$$
A: "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.
A: There is another shortcut method to find the general term $T_n$ in terms of $n$. 
We have : $T_n=T_{n-1}+3$ or $T_n-T_{n-1}=3$ implies that the difference of any two consecutive terms of the given series is constant $(d=3)$ hence it is an A.P. Now, setting $n=1$ in the recurrence relation, $$T_1=T_0+3=1+3=4$$
hence, the $n$th term of $A.P.$ $$T_n=T_1+(n-1)d=4+(n-1)\times3=3n+1$$  
