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$$\sum_{r=0}^{k-1}4^r$$ Hi, I was wondering whether anyone could explain how to work this out. I know the end result is $\frac{4^k-1}{3}$, but I don't know why or how to get there. Thank you :D

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It is a geometric sum. See in particular the section "formula" in the reference. –  Mårten W Jun 30 '13 at 8:36
    
Ahh, I get it now, thanks :D –  Elise Jun 30 '13 at 8:55

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up vote 2 down vote accepted

Multiply it by $(4-1)$. Expand without turning it into $4-1=3$. a lot of powers of $4$ will simplify except $4^k$ and $1$. Afterwards divide by $4-1=3$.

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Actually, sorry about this, but I don't understand it completely. I end up with ([4^k-1)-1]/3 instead of ([4^k]-1)/3. Thanks :D –  Elise Jun 30 '13 at 9:00
    
$4^{k-1}\times 4=4^{k}$. –  ABC Jun 30 '13 at 9:07
    
Okay, but why am I multiplying it by 4? Is the equation for this sum $\frac{a(r^n -1)}{r-1}$, with a=1, r=4 and n=k-1, if this is the case, why isn't the answer $\frac{4^(k-1) -1}{3}$ The 4 here is ment to be to the power of k-1 but I'm not too sure how to write that. Sorry for being so slow at this :S –  Elise Jun 30 '13 at 9:13
    
Woohoo, I understand now. I was thinking that n=k-1, but n=k :) it all makes sense now. –  Elise Jun 30 '13 at 14:36

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