Pumping Lemma for anb2n+1 - pumping-lemma

Question

I know how to solve pumping lemma for anbn :n>=0
But I don't understand how can I solve this example : anb2n+1 :n>=0
I tried to solve it but I am not sure that I have solved it correctly or not?Could somebody please help me here?
I can show how did I solve it. But seriously I am not sure is it correct or not. Could you please give me the correct one if I am wrong.
Question : Prove that anb2n+1 :n>=0 is not regular.
Here is my answer.
Assume L is regular. Then pumping lemma must hold. Let m be an integer in Pumping lemma.
Let w=amb2m+1 also in L. and |w|>=m
By Pumping lemma w=xyz where |xy|<=m and |y|>=1
According to pumping lemma wi=xyiz also in L for i=0,1,2,...
Let i=2 then w2=xyyz.
Let y=ak where 1<=k<=m and x=aq where 0<=q< m then z=am-q-kb2m+1
w2=xyyz = aqakakam-q-kb2m+1
= am+kb2m+1
but this is not in L for any value of 1<=k<=m
So we have contradiction with pumping lemma. so, our assumption that L is regular is wrong. So, L can not be regular.
Is this correct???
Thank you.

Answer 1

x=ak
y=aj
z=am-j-kb2m+1
Taking i=m+2 you have:
am+m+1b2m+1 ==> a2m+1b2m+1 that isn't in L.

Answer 2

Pumping Lemma is used to prove that a Language is NOT REGULAR
If A is a Regular Language, then A has a Pumping Length 'P' such
that any string 'S where |S|>=P may be divided into 3 parts S=xyz such
that the following conditions must be true:
(1) xyz & A for every i>0
(2) |y|>0
(3) |xy| <=P
To prove that a language is not Regular using PUMPING LEMMA, follow
the below steps:
(We prove using Contradiction) -> Assume that A is Regular
-> It has to have a Pumping Length (say P)
→ All strings longer than P can be pumped |S|>=P -> Now find a string
'S' in A such that |S|>P
-> Divide S into x y z
-> Show that x y z & A for some i -> Then consider all ways that S can be divided into xyz
-> Show that none of these can satisfy all the 3 pumping conditions at the same time →S cannot be Pumped == CONTRADICTION
To prove this given expression is regular by a example: Assume that
A={a^nb^2n+1|n>=0} is regular Pumping length=p
|S|=3p+1>=p |S|>=p Divide s into x y z Let p=3,and
s=aaabbbbbbb
Case1: The y is in a's part a -> x aa -> y bbbbbbb -> z xy^iz
i=2
xy^2z=>a aaaa bbbbbbb
xy^2z is not belongs to A
Case2: The y is in b's part aaa -->x bb --> y
bbbbb -->z
xy^iz
i=2
xy^2z=> aaa bbbb bbbbb
xy^2z is not belongs to A
Case3: The y is in both a and b parts a-> x
aabb ->y
bbbbb -> z
xy^iz
i=2
xy^2z=> a aabbaabb bbbbb
xy^2z is not belongs to A */ we got contradiction Hence given
expression is not regular
FOR BETTER UNDERSTANDING ABOUT PUMPING LEMMA PLEASE GO THROUGH THE LINK
https://youtu.be/Ty9tpikilAo