## Induction: have a working Proof but first step (n=1) fails

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corgan
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### Induction: have a working Proof but first step (n=1) fails

I am needing to prove the following by induction:

${n}$
${\sum } {\rm }{\rm } 2^{n -1}{\rm } =1+2+2^{2}+...+2^{n -1}=2^{n }$
${j =0}$

My proof is like so:

Assumption: $2^{(k-1)} = 2^{k}$
Proof needed: $2^{(k-1)} + 2^{(k+1-1)} = 2^{(k+1)}$
Using Assumption,
Added 2^(k+1-1) to both sides: $2^{(k-1)} + 2^{k} = 2^{k} + 2^{k}$
Factored out the 2^k on the RHS: $2^{(k-1)}+2^{k} = 2*2^{k}$
Using exponent addition rule: $2^{(k-1)}+2^{k} = 2^{(k+1)}$

Assumption = Proof, thus proven by induction.

The problem is I cannot baseline a value for n (the first required step of induction). E.g. if n=1 I get 2^0 = 2^1 which is false...

What am I missing? I am a bit confused because the algebraic proof works.

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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### Re: Induction: have a working Proof but first step (n=1) fai

corgan wrote:I am needing to prove the following by induction:

${n}$
${\sum } {\rm }{\rm } 2^{n -1}{\rm } =1+2+2^{2}+...+2^{n -1}=2^{n }$
${j =0}$

Since every term will be even except for the first term (which is 1), then the sum must be odd. Therefore, the sum cannot be a power of 2.

Also, the counter is given as being "j", so the terms should have powers in terms of j, not in terms of n. Also, if the first counter-value is 0, then the power cannot be in the form given above, as this would create a fraction for the first term.

Something is clearly wrong with the problem statement. Please consult with your instructor regarding corrections. Thank you!