My A-level textbook has this example:
Given that $u_{n+1}=3u_n+4, u_1=1$ and $n\ge1$, prove by induction that $u_n=3^n-2$
The solution presented is as follows:
$n=1;\,u_1=3^1-2=1$ as given.
$n=2;\,u_2=3^2-2=7$, from the general statement and $u_2=3u_1+4=3(1)+4=7$, from the recurrence relation. So $u_n$ is true for $n=1$ and is also true for $n=2$.
Assume that for $n=k, u_k=3^k-2$ is true for $k\in Z^+$. Then
$\begin{equation}\begin{aligned} u_{k+1}&= 3u_k + 4 \\ &= 3(3^k-2) + 4 \\ &= 3^{k+1}-6+4\\ &= 3^{k+1}-2 \end{aligned}\end{equation}$
Therefore, the general statement, $u_n$ is true when $n=k+1.$ If $u_n$ is true when $n=k$, then it has shown that $u_n$ is also true when $n=k+1.$ As $u_n$ is true for $n=1$ and $n=2 \;$ then $u_n$ is true for all $n\ge1$ and $n\in Z^+$ by mathematical induction.
Is the basis for $n=2$ necessary? The assumption steps made use of the fact that if $u_n$ is true for $n=k$, then $u_n$ is true for $n=k+1$. Therefore the basis for $n=1$ is sufficient, in my opinion, cognizant of the fact that $u_1$ is a given value and not obtained from the recurrence formula. Am I wrong? The textbook did explain this though (the rational of which I did not quite get)
** In the basis step of the proof, the general statement was checked for $n=1$ and $n=2$. This is because the first application of the recurrence formula $u_{n+1} = 3u_n+4$ yield $u_2$ by using the first given term $u_1=1$ ***
There are situations where your argument for A(n) => A(n+1) is not valid say for n = 3, only for n >= 4. In that situation you would need to show A(1) to A(4).
In your case, the induction step is correct for n = 1 so no need to show A(2).