Question Regarding an Induction Proof from (Principle of Mathematical Induction, Chapter 4 in the NCERT Textbook) This is a small doubt regarding a question from the NCERT exemplar questions on the chapter Principle of Mathematical Induction (actually it's a deleted chapter in the sense that it's been eliminated from the syllabus that the CBSE had devised for the current year, alongside Binomial Theorem and Mathematical Reasoning). I tried to do this one question and would like to verify if the methodology I used was right.

A series starts with its first term $b_0 = 5$ and progresses as $b_k = b_{k-1}+4$. Prove that the elements of the series are of the form $5 + 4n$ for all natural number $n$.

I did as follows:
Let $S = \lbrace b_n:b_0 = 5, b_n = b_{n-1} + 4, n \in \mathbb{N}\rbrace$
Let $P(k) = 5+4k \in S$
Base case: We see that $P(1)$ is true ($9 \in S$) and also $P(2)$ also holds true ($13 \in S$)
Inductive step: Let's assume that this holds for all $1\leq n \leq k \implies P(k)$ is true and hence $5 + 4k \in S$, so we can safely assume $b_k = 5+4k$
We see, $5 + 4(k+1) = \underline{5+4k} + 4 = b_k + 4$ which essentially turns out to be $b_{k+1}$  and thus we see, $5+4(k+1) = b_{k+1} \in S$ and hence $P(k+1)$ is also true.
Thus we see that $$b_n = 5 + 4n \space \forall\space  n \in \mathbb{N}$$.
I haven't tried this sort of a proof before so I'd like to learn from you what is wrong with the proof, what assumptions are actually not worth doing or are dangerous to add fallacies/loopholes to the proof.
 A: Critiques.
My first critique is were the Proposition states "the elements of the series are of the form 5+4n for all natural number n" and you state "Let $P(n) = 5+4k \in S$"
I don't think the wording of the proposition is very clear.  ""the elements of the series are of the form 5+4n"  would mean that for every $b_k$ there is an $n$ so that $b_k = 5+4n$.  But the clause "for all natural number n" muddles it.  Are the saying for every natural $n$ there is $k$ so that $5+4n=b_k$?  So are the saying that the set $\{b_k\}$ and the set of all $\{5+4n|n\in \mathbb N\}$ are the same set (possibly with some doubling up)?
It'd have been a lot more straightforword (and easier to prove) if the proposition had been that stronger (and still true) statment that for every $n\in \mathbb N$ we have $b_n = 5+ 4n$.  But the proposition didn't say that.  As best I can intepret it the proposition says "For every $n \in \mathbb N$ then $5+4n \in S$ and for ever $b_k\in S$ then $b_k = 5+4n$ for some $n\in \mathbb N$".  That is two things to prove.
(This is the books fault not yours.)
Your statement:  $P(n) = 5+4k \in S$ is hard to parse.  It take it you mean: The statement $P(n)$ is $5+4k \in S$.  But the variables $n$ and $k$ don't agree.  I'll take it you mean for all $n \in \mathbb N$ that $5+4n\in S$ so $P(N):: 5+4n \in S$ would be what you mean.
But this is only half the statement.  Proving $P(n)$ for all $n$ would show us $\{5+4n|n\in \mathbb N\}\subset S$.
The other half would be $Q(n):: b_n = 5+4k$ for some natural number $k$ and proving that will prove $S \subset \{5+4n|n\in \mathbb N\}$.  Together that proves $S = \{5+4n|n\in \mathbb N\}$.
But.... this is ridiculous.....
We should just state:  We are going to prove $P(n)::b_n = 5 + 4n$ for all $n\in \mathbb N$.  That is what you proved in the end anyway....
....
Now your proof:
"We see that P(1) is true (9∈S) and also P(2) also holds true (13∈S)"
Okay, maybe we should point out that $5 + 4\cdot 1 = 9$ and $9\in S$ because $b_1 = b_0 + 4 = 5 + 4=9$; and that $5+4\cdot 2 = 13$ and $13\in S$ becase $b_2 = b_1+4 = 9 + 4 = 13$-- just to be clear, but yes, that is correct.
"Inductive step: Let's assume that this holds for all 1≤n≤k⟹P(k) is true and hence 5+4k∈S, so we can safely assume $b_k=5+4k$"
Ah, see!  But your proposition was never that $5 + 4k = b_k$.  Your proposition was that $5 + 4k\in S$.  Maybe $5 + 4\cdot 87 = b_{153}$.....  But you know what.... I blame your text, not you,  it's pretty clear that proving $b_k= 5+4k$ is a stronger and true statement.
(FWIW.... if I wanted to prove $5+4k \in S \implies 5+4(k+1) \in S$ without assuming that $5+4k = b_k$ we could do this:  If $5+4k \in S$ then there is a $b_m$ so that $b_m = 5+4k$.  Therefore $b_{m+1} = 5+4k +4 = 5+4(k+1)$.  So $5+4(k+1) = b_{m+1} \in S$.... but that's .... kinda silly, isn't it... back to your proof....)
"We see, 5+4(k+1)=5+4k––––––+4=bk+4 which essentially turns out to be bk+1 and thus we see, 5+4(k+1)=bk+1∈S and hence P(k+1) is also true."
Yes, that's great.  Just fine.
"Thus we see that
bn=5+4n ∀ n∈N"
Okay.... If our proposition from the start had been:  $P(n):: b_n = 5+4n$ and we had done

Induction:  $k=0$.   $b_0 = 5 = 5+4\cdot 0$.  $b_1 = b_0 + 4 = 5+4 = 9 = 5 + 4\cdot 1$.

We'd be just fine.
===
FWIW if I wanted to do exactly what the book said and prove:  $5n+4\in S$ for all $n\in \mathbb N$ and for all $b_n \in S$ then $b_n = 5 +4k$ for some $k \in \mathbb N$, I'd do
$P(n):: 5n+4 \in S$.
Base Case:  $n= 0$ then $5+4\cdot 0 =5 = b_0$.
Induction step:  $P(k)\implies P(k+1)$.  Suppose that $5 + 4k =b_m$ for some $b_m \in S$.  Then $b_{m+1}= b_m + 4 = 5+4k + 4 = 5 + 4(k+1)$.  So $5+4(k+1)=b_{m+1} \in S$. So $P(k+1)$ is varified.
$Q(n)::$ every $b_n = 5 + 4m$ for some $m \in \mathbb N$.
Base Case:: $n = 0$.  $b_0 = 5= 5+4\cdot 0$.
Induction Steep:: $Q(k)\implies Q(k+1)$
Suppose $b_k = 5 + 4m$.  Then $b_{k+1} = 5 + 4m + 4 = 5+4(m+1)$ and $m+1 \in \mathbb N$.  So $Q(k+1)$ is varified.
.....
But that was silly, wasn't it.
A: There are no loopholes in this proof. It's absolutely fine. The key thing to observe before starting with this proof (and you have observed that) is to note that
\begin{align*}
b_n&=b_{n-1}+4\\
&=b_{n-2}+4+4\\
&=b_{n-2}+2\times 4\\
&=b_{n-3}+3\times 4\\
&\;\;\;\;\cdots\\
&\;\;\;\;\cdots\\
&\;\;\;\;\cdots\\
&=b_{n-n}+n\times 4\\
&=5+4n
\end{align*}
