Find the close form

Question

$a_0 = 0$ and, for $n > 0$, $a_n =a_{n-1} + 3$ be a recurrence relation. The close form of $a_n$ is:

A. $3(n-1)$

B. $3n$

C. $3(n+1)$

D. $n^2$

I guess is B but anyone could explain it for me please? Thanks

Answer 1

It is an arithmetic recurrence so its general term is $3n+a_0=3n$.

You can prove it as follows:

$a_n=a_{n-1}+3$

$a_{n-1}=a_{n-2}+3$

...

$a_1=a_0+3$

Now sum all the previous equations member by member. You obtain:

$a_n=a_0+3n=3n$