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$2^0+2^1+2^2+...+2^n$ for $n ∈ \mathbb{N} \cup \{0\}$.

I made a conjecture that this is $2^{n+1} - 1$. Now I have to prove it by induction.

I tested the base case where it's equal to zero, and it worked.

Then I did "assume $k∈ \mathbb{N} \cup \{0\}$ and $2^0+2^1+2^2+...+2^k$". I got marked down for saying "Let $n=k+1$", and did a bunch of math for it to become $2^{k+2} -1$ and I was apparently wrong >_>?

Martin Sleziak
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Test
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    http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series – copper.hat Dec 03 '14 at 22:54
  • What is the question? – copper.hat Dec 03 '14 at 22:54
  • How do I prove 2^0+2^1+2^2+...+2^n for n ∈ N∪{0} = 2^(n+1) -1

    I mean, I wrote exactly what that link says and I got marked down. Is there a way to post a picture of what I wrote?

     – Test Dec 03 '14 at 22:56
  • http://imgur.com/mI2SkbF Here's a link to my homework and what I got wrong. I really can't figure it out. – Test Dec 03 '14 at 22:58
  • @Test The marking goes for the wording chosen. You should have left out the "Let $n=k$" and "Let $n=k+1$". The rest is okay. – AlexR Dec 03 '14 at 23:01
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    @Test You assumed what you were trying to prove in the first line after "Let $n=k+1$." If you had left off the quantity on the RHS of the equals sign in that centered line, you would have gotten full (or nearly full) credit. – apnorton Dec 03 '14 at 23:03
  • @AlexR Um... Are you sure? (I want to make sure I'm reading it right.) – apnorton Dec 03 '14 at 23:03
  • @anorton At least I would have accepted that. The computations are correct and if you write "Let $k\in \mathbb N_0$ and $\sum_{i=0}^k 2^i = 2^{k+1}$. Then (...) $\Rightarrow \sum_{i=0}^{k+1} 2^i = 2^{k+2} - 1$." It's fine to me. – AlexR Dec 03 '14 at 23:06
  • @AlexR Ok. I see. – apnorton Dec 03 '14 at 23:07
  • @Test Your corrector probably wants you to learn how to write accurate proofs, or he had a really bad day. – AlexR Dec 03 '14 at 23:07
  • He demands very accurate proofs. but he gave me 25% credit. he said that "Let n=k+1" is WRONG under the guidelines of "Incorrect assumption", and then said that I have to derive 2^(k=1) -1 case from here. So I don't really know what to correct or what to do. – Test Dec 03 '14 at 23:10
  • I'll second what apnorton said. Disregarding the incorrect use of the word "let", for the $n=k+1$ case, the first thing you wrote was that it equaled $2^{k+2}-1$. That's wrong! You can't just say it equals that, you must show it. While you continued to demonstrate this, the damage was done. It's a step that is not logically in order. – zahbaz Nov 02 '15 at 09:41

1 Answers1

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You have to prove it on both sides, but you can make the assumption that you prove it for earlier cases.

So to prove for $n+1$ you may assume it holds for $n$.

Pieter21
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  • I know, but what was the mistake I made? I set assume it's true for n=k, and try to prove it for k+1, which I did, but I got it wrong. here's a pic of my homework: http://www.imgur.com/mI2SkbF – Test Dec 03 '14 at 23:03
  • I think your proof is correct though unclear and the corrector may have misunderstood, or is deducting points for unclarity. The 'Let' stuff is unclear because we're not doing assignments. Symbol rewrite is also not necessary and unclear. If instead of writing 'Let n = k + 1' you'd say, Let's calculate for n + 1, and not already put what you need to prove on RHS, it would be OK. – Pieter21 Dec 04 '14 at 11:00