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I'm having trouble to find context free grammars which generate following languages:

  1. $L_1 = \{a^nb^{2n}∣n\ge1\}$

  2. $L_2 = \{a^pb^{p+q}a^q∣p,q\ge1\}$

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    Welcome to Math.SE! Questions which elaborate on what the asker is having difficult with or confused about are the most likely to receive helpful answers. – 2017-01-10

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HINT: For $L_1$ you need a very small modification of a context-free grammar for $\{a^nb^n:n\ge 1\}$, and I suspect that you’ve seen such a grammar: it has just two productions, $S\to aSb$ and $S\to ab$. That same grammar contains the essential idea needed for a grammar for $L_2$ as well. Start with a production $S\to XY$, let $X$ produce strings of the form $a^pb^p$, and let $Y$ do the rest.

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    So for L1 is it just two productions with S → aSbb and S → abb? – 2017-01-10
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    @Tes: That’s right. – 2017-01-10
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    And for L2 would this be correct: S → XY, X → aXb, X → ab, Y → bYa, Y → ba? – 2017-01-10
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    @Tes: Yes, exactly. – 2017-01-10
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    wow i understand it now, thank you so much! – 2017-01-10
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    @Tes: You’re welcome! – 2017-01-10