I am having trouble understanding how to apply Pumping Lemma to show a Language is not regular.
If the alphabet is $\Sigma = \{a, b\}$ and the language is $L = \{a^{i + 1} b^{4i + 2} \mid i \in \mathbb N\}$, how do I prove the language is not regular using pumping Lemma?
Also what is a pumping length, how do i know to assign a pumping length to the language stated above?
Since I see questions about the pumping lemma on this site quite often, I decided to write a bit of a longer answer hoping that it helps people "get" the PL rather than just treating it as a "plug-n-chug" tool given to us by the gods of math.
I think the best way to go about it is to basically (re-)derive the lemma. The idea is that if some language $L$ is regular, then there exists some DFA $A$ such that $A$ recognizes $L$. Now say that $A$ has $m$ states and suppose that we have some $w \in L$ with of length $n > m$ (note that any infinite $L$ must contain such a word) and consider the run of $A$ on $w$.
As $w$ has more characters than $A$ has states, there must be some state $q$ which occurs at least twice in $A$'s run on $w$, i.e. if we draw out the run through the graphic representation of $A$ we find some loop. In more formal terms, this means that there must be some substring $s$ of $w$ -- so $w$ can be written as $w = rst$ where $r$ and $t$ are remaining parts of $w$ left and right of $s$ -- such that $A$ is in state $q$ on the both the first and last symbol of $s$.
At this point, we need to understand the D part of DFA: $A$ is deterministic, so given that we start from the initial state on any input of the form $r\Sigma^\ast$ (any string starting with the substring $r$) we will always end up at the state $q$ and similarly, starting from $q$ on the word $s$ we always end up at $q$ again and finally, starting on $q$ given the word $t$ we always end up at the same (accepting) state. I like to think of such deterministic machines as music boxes; if you start the at the same point they end up playing the same melody.
This means that we can now create new words which $A$ must also accept by making $A$ go into state $q$ and presenting it with $t$, and the easiest way to do so is by taking the known string $r$ to get it to the state $q$ initially, then repeating $s$ some $k \geq 0$ times (which, because we start at $q$ now) will always get us back to $q$ and then appending $t$ to make $A$ go into an accepting state. Hence these words are of the form $w_k = rs^kt$ for some $k \geq 0$.
Taking a step back, we have now basically argued that $A$ cannot distinguish between $w$ and $w_k$; the result of the computation will always be the same. This is the essence of how the pumping lemma works.
To apply the pumping lemma we must now find such $w$ and $k$ where $w \in L$ and $w_k \notin L$. Usually one will not be able to give $w$ directly as the size of the automaton $A$ is unknown (in fact we are trying to argue that such an $A$ does not exist, so yeah). In most cases, this boils down to taking a "large" string in $L$ such that the middle part of said string is big enough for us to find some part that if repeated would give us the $w_k$ we need.
Let us get specific now. Our language is $L = \{a^{i + 1} b^{4i + 2} \mid i \in \mathbb N\}$. We assume that $L$ is regular and thus admits a DFA $A$ of size $m$. If we consider the string $w = a^{10m + 1} b^{40m + 2} \in L$, we find that a state repetition must already occur in the $a$-part of $w$ as $a^{10m + 1}$ has more symbols than $A$ has states. Hence there must be a decomposition $w = rst$ where $r = a^{\ell}$, $s = a^k$ and $t = a^{10m + 1 - \ell - k} b^{40m + 2}$ with $0 \leq \ell < m$ and $0 < k \leq m$ (Can you figure out why the bounds on these variables are this way? This is a good test to see if you are confident with the first part, I think).
If we now consider the word $$ w_2 = rs^2t = a^\ell (a^k)^2 a^{10m + 1 - \ell - k} b^{40m + 2} $$ and unpack it to $$ a^{m + k + 1} b^{40m + 2} $$ we see that $w_2$ is not in $L$. It follows that $A$ does not recognize $L$ as we assumed that $A$ accepts $w \in L$ and then argued that it must also accept $w_2 \notin L$. As the choice of $A$ was arbitrary we find that no DFA can recognize $L$, which means that $L$ is not regular.
Assume the language is regular, i.e., there is an automaton that recognize the language with a finite set of states. Therefore there are $i \neq j$ such that processing $a^{i +1}$ and $a^{j + 1}$ reach the same state X. Processing $4i + 2$ b's in state X always ends up in the same state Y. State Y must be accepting because $a^{i + 1} b^{4i + 2}$ is in the language. At the same time it must not be an accepting state, because $a^{j+ 1} b^{4i + 2}$ is not in the language when $i \neq j$. Contradiction, so the language is not regular.
