Pumping Lemma for anb2n+1 -

i know how solve pumping lemma aⁿbⁿ :n>=0 don't understand how can solve example : aⁿb²ⁿ⁺¹ :n>=0

i tried solve not sure have solved correctly or not?could please me here?

i can show how did solve it. not sure correct or not. please give me correct 1 if wrong.

question : prove aⁿb²ⁿ⁺¹ :n>=0 not regular.

here answer.

assume l regular. pumping lemma must hold. let m integer in pumping lemma.

let w=a^mb^2m+1 in l. , |w|>=m

by pumping lemma w=xyz |xy|<=m , |y|>=1

according pumping lemma w_i=xyⁱz in l i=0,1,2,...

let i=2 w₂=xyyz.

let y=a^k 1<=k<=m , x=a^q 0<=q< m z=a^m-q-kb^2m+1

w₂=xyyz = a^qa^ka^ka^m-q-kb^2m+1

= a^m+kb^2m+1

but not in l value of 1<=k<=m

so have contradiction pumping lemma. so, our assumption l regular wrong. so, l can not regular.

is correct???

thank you.

x=a^k
y=a^j
z=a^m-j-kb^2m+1

taking i=m+2 have:

a^m+m+1b^2m+1 ==> a^2m+1b^2m+1 isn't in l.