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Alright I have this problem, Prove by induction $2^1+2^2+2^3+2^4+ \cdots +2^n=2(2^n-1)$

Now I've done this so far:

Base case $n=1$:

$$2^1 = 2$$

$$2(2^1-1)=2(2-1)=2(1)=2 .$$

Assume for $k$, prove for $k+1$:

$$ \begin{align*} 2^1+2^2+2^3+2^4+ \cdots +2^k+2^{k+1} & =2(2^k-1)+2^{k+1} \\ &=2^{k+1}+2^{k+1}-2 \end{align*} $$

Now the trouble I'm running into is that I don't know how to continue from here, I know that I need to show that somehow $2^{k+1}+2^{k+1}-2 = 2(2^{k+1}-1)$.

Is there something that I would be missing with the rules of exponents, or maybe I just made a mistake and I'm doing something wrong that I don't recognize?

Any help would be appreciated, thanks.

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We have $2^{k+1} + 2^{k+1} = 2^{k+1} (1+1)$. I'll let you conclude. –  Joel Cohen Dec 3 '11 at 23:54
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$2^{k+1}+2^{k+1}=2\cdot 2^{k+1}$ –  Brian M. Scott Dec 3 '11 at 23:54
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You're almost finished, what's the problem? $2^{k+1}+2^{k+1}-2 = 2\cdot 2^{k+1}-2\cdot 1 = 2\cdot(2^{k+1}-1)$ –  Martin Sleziak Dec 3 '11 at 23:55
    
Duplicate of this answer : math.stackexchange.com/questions/86838/… –  Patrick Da Silva Dec 4 '11 at 1:20
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2 Answers

$2^{k+1}+2^{k+1}-2 = 2\cdot 2^{k+1}-2 = 2(2^{k+1}-1)$

In first equality I use $a + a = 2a$, in second equality I factor out 2 (distributivity).

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hahaha thank you so much, when you put it like that I can see exactly what I was missing. –  Latency Dec 4 '11 at 0:01
    
If this question completed your request, you should accept it. –  Dhaivat Pandya Dec 4 '11 at 2:00
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In my duplicate that I've posted in the comments, there are many ways to do this that are "not" done by induction. One good exercise for you would be to prove the exact same thing but replace $2$ with $b$ and show that $$ b^1 + b^2 + \dots + b^n = b \frac{b^n -1}{b-1} $$ for any real number $b \neq 1$. This expression comes up very often in mathematics so if you plan on doing maths a little in your life it is important that you recognize those.

Hope that helps,

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