title: Pumping Lemma for anb2n+1
body: <p>I know how to solve pumping lemma for a<sup>n</sup>b<sup>n</sup> :n&gt;=0
  But I don't understand how can I solve this example : a<sup>n</sup>b<sup>2n+1</sup> :n&gt;=0</p>
  <p>I tried to solve it but I am not sure that I have solved it correctly or not?Could somebody please help me here?</p>
  <p>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.</p>
  <p><strong>Question :</strong> Prove that a<sup>n</sup>b<sup>2n+1</sup> :n&gt;=0 is not regular.</p>
  <p>Here is my answer.</p>
  <blockquote>
  <p>Assume L is regular. Then pumping lemma must hold. Let m be an integer in Pumping lemma.</p>
  <p>Let w=a<sup>m</sup>b<sup>2m+1</sup> also in L. and |w|&gt;=m</p>
  <p>By Pumping lemma w=xyz where |xy|&lt;=m and |y|&gt;=1</p>
  <p>According to pumping lemma w<sub>i</sub>=xy<sup>i</sup>z also in L for i=0,1,2,...</p>
  <p>Let i=2 then w<sub>2</sub>=xyyz.</p>
  <p>Let y=a<sup>k</sup> where 1&lt;=k&lt;=m and x=a<sup>q</sup> where 0&lt;=q&lt; m then z=a<sup>m-q-k</sup>b<sup>2m+1</sup></p>
  <p>w<sub>2</sub>=xyyz = a<sup>q</sup>a<sup>k</sup>a<sup>k</sup>a<sup>m-q-k</sup>b<sup>2m+1</sup></p>
  <p>= a<sup>m+k</sup>b<sup>2m+1</sup></p>
  <p>but this is not in L for any value of 1&lt;=k&lt;=m</p>
  <p>So we have contradiction with pumping lemma. so, our assumption that L is regular is wrong. So, L can not be regular.</p>
  </blockquote>
  <p>Is this correct???</p>
  <p>Thank you.</p>