How would I show that Tn=3^n + 2 is a solution to the recurrence?

Question

Would anyone be able to help me or give me some advice on the following problem:

Consider the recurrence with $T_0 = 3$ and $T_{n+1} = 3T_n - 4$ for all $n \in \mathbb{N}$.

How would I show that $T_n=3^n + 2$ is a solution to the recurrence?

Answer 1

$$T_{n+1}-1=3(T_n-1)=\cdots=3^r(T_{n+1-r}-1)$$ where integer $r\ge0$

Answer 2

It is given that $T(n+1)=3T(n)-4$ for all $n \in N.$
Clearly $T(0)=3^0+2=1.$
Suppose $T(k)=3^k+2$ for some $k \in N.$
Then $T(k+1)=3(3^k+2)-4=3^{k+1}+2.$
Heance, by mathematical induction $T(n)=3^n+2, \forall n \in N.$

Answer 3

If you simply want to verify, just substitute, $$T_{n+1}=3T_n-4$$ to $$(3^{n+1}+2)=3(3^n+2)-4. $$ And we see it works. :-)