What does \(1 + 2 + 4 + 8 + 16 + 32 + ...\) equal?

You could rewrite the equation as \((2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)\)

This is then written as \((2 + 4 + 8 + 16 + 32 + ...) - 1 - 2 - 4 - 8 - 16 - 32 - ...\).

Which results in \(1 + 2 + 4 + 8 + 16 + 32 + ...= -1\)!!!.

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## Comments

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TopNewestThis is an infinite geometric series which has ratio \(r=2>1\). So this doesn't converge, and as the result we get an non-sense answer.

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I do think this answer is a bit weird, but the method seems correct, since it is an

infiniteseries! and does not end.Log in to reply

I'm sorry but you can't just use the distributive laws when dealing with infinitely many terms. A simple proof that this is wrong is that a sum diverges when its partial sums diverges. Clearly the partial sums are \(2^n-1\) which approaches to infinity as \(n\) does, therefore the sum does not exist and is undefined.

Or instead you are talking about something much more complicated like in a different number system, like 2-adic numbers. :P

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Damn, we communicated before Schol of Excellence???

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It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important

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There is no largest term.

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ahh I like these fallacious proofs - they really test your observation. I think the reason why you got -1 was because the factorization only works if the expression is finite.

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these laws are not valid when we are dealing with infinite sums.

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I had a similar derivation which I am not so sure about. My first note on Summations

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Just an illusion.....,

Confine up to 5 or 6 terms and try to evaluate..

## InitiallyTrolled me too..,

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Nice logic!!

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I do not see where \(-1\) came from.

EDIT: For a proof of where this summation went wrong, check my comment down here.

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Remove the brackets and you get 2+4+8+16+32+...-1-2-4-8-16-32. The +2 and -2 cancel each other out, so does +4 and -4, +8 and -8, etc. until all that's left is -1

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This is obviously false, we still have a term in the first sequence missing.

If the RHS

ends, let's say, in \(2^n\), the LHS will end in \(2^{n+1}\). This leads us that the \( \Sigma = 2^{n+1} - 1\), which is true.But we cannot evaluate a limit for \(n \rightarrow \infty\), because \(\Sigma\) would tend also to infinity.

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does notend. The point of infinite sequences is that they do not end. Even the proof for \(9.99999\ldots=10\) requires the \(9\)s to never end.Log in to reply

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