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How can I prove the below?

$$2^{pq} - 1 = (2^p - 1)\left(\sum^{q-1}_{i=0} 2^{pi}\right)$$ for two natural numbers $p, q$

It looks like I need proof by induction? But how? There's two variables?

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2 Answers 2

up vote 2 down vote accepted

Setting $a=2^p$ then $$ 2^{pq}-1=a^q-1=(a-1)\sum_{i=0}^{q-1}a^i=(2^p-1)\sum_{i=0}^{q-1}(2^p)^i=(2^p-1)\sum_{i=0}^{q-1}2^{pi}. $$

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For the summation, can't we take 2^p out of the summation and use sum of a geometric progression formula?

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If this is intended as a question, it should be a comment, and not an answer. You have 84 rep so you should be able to comment, IIRC. –  Pedro Tamaroff Aug 25 '12 at 1:44
    
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Sasha Sep 2 '12 at 3:39

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