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This is true? $ 1+1 + 2+ 2^2 +2^3 + \cdots + 2^{30}= 2^{31}$, How to prove?

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    $\begingroup$ Do you know anything about geometric series? $\endgroup$
    – user296602
    Commented Feb 13, 2016 at 18:44
  • $\begingroup$ no i dont any thing $\endgroup$
    – Soroush
    Commented Feb 13, 2016 at 18:45
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    $\begingroup$ Well, then get a calculator.... $\endgroup$
    – user296602
    Commented Feb 13, 2016 at 18:45
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    $\begingroup$ @MichaelHardy I think an elementary-algebra tag is way too vague, especially since this question fits squarely in the scope of algebra-precalculus. $\endgroup$
    – user296602
    Commented Feb 13, 2016 at 19:07
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    $\begingroup$ @MichaelHardy So in your opinion, the fight for changing what "algebra precalculus" means starts with polluting math.SE tags? $\endgroup$ Commented Feb 14, 2016 at 8:09

4 Answers 4

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First write:

$$ \sum_{k=0}^{n+1} x^k = 1 + x\sum_{k=0}^{n} x^k \\ \Rightarrow (1-x)\sum_{k=0}^{n}x^k = 1 - x^{n+1} \\ \Rightarrow \sum_{k=0}^{n}x^k = \frac{1 - x^{n+1}}{1 - x} $$

Now your sum is just

$$ 1 + \sum_{k=0}^{30} 2^k = 1 + \frac{1 - 2^{31}}{1- 2}\\ = 1 + 2^{31} - 1 \\ = 2^{31} $$

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Yes, it is true. Try it for some small values to convince yourself. To prove it, note that the sum after the first term is a geometric series and sum it. For the specific case, you can just compute both sides. That is a fine proof.

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If you don't know anything about geometric series, $2^n-1=1+2+\cdots+2^{n-1}$ can be easily proved by induction.

It's clearly true for $n=1$. Suppose it's true for some $n=k$ and consider $n=k+1$.

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

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For computing the geometric sum $S=1+x+x^2+\cdots+x^n$ compare it to $$ x\cdot S=x+x^2+x^3+\cdots+x^{n+1} $$ to find that the difference is $$ (x-1)\cdot S=x^{n+1}-1 $$ Now set $x=2$…

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